ML Aggarwal Solution Class 10 Chapter 18 Trigonometric Identities Test

Test

Question 1

(i) If θ is an acute angle and cosec θ = √5 find the value of cot θ – cos θ.

(ii) If θ is an acute angle and $\tan \theta=\frac{8}{15}$ find the value of sec θ + cosec θ.

Sol :

(i) θ is an acute angle.

cosec θ = √5

$\therefore \sin \theta=\frac{1}{\sqrt{5}}$

$\cos \theta=\sqrt{1-\sin ^{2} \theta}$

$=\sqrt{1-\left(\frac{1}{\sqrt{5}}\right)^{2}}=\sqrt{1-\frac{1}{5}}$

$=\sqrt{\frac{5-1}{5}}=\sqrt{\frac{4}{5}}=\frac{2}{\sqrt{5}}$

Now, cotθ-cosθ$=\frac{\cos \theta}{\sin \theta}-\cos \theta$

$=\frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}}-\frac{2}{\sqrt{5}}=\frac{2}{1}-\frac{2}{\sqrt{5}}$

$=2\left(1-\frac{1}{\sqrt{5}}\right)=\frac{2(\sqrt{5}-1)}{\sqrt{5}}$

(ii) $\tan \theta=\frac{8}{15}$

In the figure,

$\tan \theta=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{8}{15}$

$\therefore \mathrm{BC}=8, \mathrm{AB}=15$

$\therefore \mathrm{AC}=\sqrt{\mathrm{AB}^{2}+\mathrm{BC}^{2}}=\sqrt{5^{2}+8^{2}}$

$=\sqrt{225+64}=\sqrt{289}=17$

$\therefore \sec \theta=\frac{A C}{A B}=\frac{17}{15}$

$\operatorname{cosec} \theta=\frac{A C}{B C}=\frac{17}{8}$

$\therefore \sec \theta+\operatorname{cosec} \theta=\frac{17}{15}+\frac{17}{8}$

$=\frac{136+255}{120}=\frac{391}{120}=3 \frac{31}{120}$

Question 2

Evaluate the following:

(i) $2 \times\left(\frac{\cos ^{2} 20^{\circ}+\cos ^{2} 70^{\circ}}{\sin ^{2} 25^{\circ}+\sin ^{2} 65^{\circ}}\right)$– tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°

(ii) $\frac{\sec 29^{\circ}}{\operatorname{cosec} 61^{\circ}}$+ 2 cot 8° cot 17° cot 45° cot 73°0 cot 82° – 3(sin2 38° + sin2 52°)

(iii) $\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}$+ $sin^2$ 63° + cos 63° sin 27°

Sol :

(i) $2 \times\left(\frac{\cos ^{2} 20^{\circ}+\cos ^{2} 70^{\circ}}{\sin ^{2} 25^{\circ}+\sin ^{2} 65^{\circ}}\right)$– tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°

$=2\left[\frac{\cos ^{2}\left(90^{\circ}-70^{\circ}\right)+\cos ^{2} 70^{\circ}}{\sin ^{2} 25^{\circ}+\sin ^{2}\left(90^{\circ}-25^{\circ}\right)}\right]-\tan 45^{\circ}$$+\tan 13^{\circ} \tan 77^{\circ} \tan 23^{\circ} \tan 67^{\circ} \tan 30^{\circ}$

$=2\left[\frac{\sin ^{2} 70^{\circ}+\cos ^{2} 70}{\sin ^{2} 25^{\circ}+\cos ^{2} 25^{\circ}}\right]-1+\tan 13^{\circ} \tan$ $-1+\tan 13^{\circ} \tan$

$\left(90^{\circ}-13^{\circ}\right) \tan 23^{\circ} \tan \left(90^{\circ}-23^{\circ}\right) \times \frac{1}{\sqrt{3}}$

$=2\left(\frac{1}{1}\right)-1+\tan 13 \cot 13^{\circ} \tan 23^{\circ} \cot 23^{\circ}\times \frac{1}{\sqrt{3}}$

$=2-1+1 \times 1 \times \frac{1}{\sqrt{3}}$

$=2-1+\frac{1}{\sqrt{3}}^{\cdot}$

$=1+\frac{1}{\sqrt{3}}=\frac{\sqrt{3}+1}{\sqrt{3}}$

$=\frac{(\sqrt{3}+1) \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

$=\frac{3+\sqrt{3}}{3}$

(ii) $\frac{\sec 29^{\circ}}{\operatorname{cosec} 61^{\circ}}+2 \cot 8^{\circ} \cot 17^{\circ} \cot 45^{\circ} \cot$ 

$73^{\circ} \cot 82^{\circ}-3\left(\sin ^{2} 38^{\circ}+\sin ^{2} 52^{\circ}\right)$

$=\frac{\sec 29^{\circ}}{\operatorname{cosec}\left(90^{\circ}-29^{\circ}\right)}+2 \cot 8^{\circ} \times \cot \left(90^{\circ}-\right.\left.8^{\circ}\right) \times \cot 17^{\circ} \times \cot \left(90^{\circ}-17^{\circ}\right) \cot 45^{\circ}-3\left[\sin ^{2} 38^{\circ}+\sin ^{2}\left(90^{\circ}-38^{\circ}\right)\right]$

$=\frac{\sec 29^{\circ}}{\sec 29^{\circ}}+2 \cot 8^{\circ} \tan 8^{\circ} \times \cot 17^{\circ} \tan 17^{\circ} \times 1-3\left(\sin ^{2} 38^{\circ}+\cos ^{2} 38^{\circ}\right)$

$=1+2 \times 1 \times 1 \times 1-3 \times 1=1+2-3=0$

(iii) $\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}$

$=\frac{\sin ^{2} 22^{\circ}+\sin ^{2}\left(90^{\circ}-22^{\circ}\right)}{\cos ^{2} 22^{\circ}+\cos ^{2}\left(90^{\circ}-22^{\circ}\right)}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin \left(90^{\circ}-63^{\circ}\right)$

$=\frac{\sin ^{2} 22^{\circ}+\cos ^{2} 22^{\circ}}{\cos ^{2} 22^{\circ}+\sin ^{2} 22^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ}\times \cos 63^{\circ}$

$=\frac{1}{1}+\left(\sin ^{2} 63^{\circ}+\cos ^{2} 63^{\circ}\right)$

=1+1=2

Question 3

If $\frac{4}{3}\left(\sec ^259^{\circ}-\cot ^231^{\circ}\right)-\frac{2}{2} \sin 90^{\circ}+3 \tan^256^{\circ} \tan ^234^{\circ}=\frac{x}{2}$then find the value of x.

Sol :

Given

$\frac{4}{3}\left(\sec ^259^{\circ}-\cot ^231^{\circ}\right)-\frac{2}{2} \sin 90^{\circ}+3 \tan^256^{\circ} \tan ^234^{\circ}=\frac{x}{2}$

$\Rightarrow \frac{4}{3}\left[\sec ^{2} 59^{\circ}-\cot ^{2}\left(90^{\circ}-59^{\circ}\right)\right]-\frac{2}{3} \sin 90^{\circ}+3 \tan ^{2} 56^{\circ} \tan ^{2}\left(90^{\circ}-56^{\circ}\right)=\frac{x}{3}$

$=\frac{4}{3}\left[\sec ^{2} 59^{\circ}-\tan ^{2} 59^{\circ}\right]-\frac{2}{3} \times 1+3\tan ^{2} 56^{\circ} \cot ^{2} 56^{\circ}=\frac{x}{3}$

$=\frac{4}{3} \times 1-\frac{2}{3}+3 \times 1=\frac{x}{3}$

$\Rightarrow \frac{4}{3}-\frac{2}{3}+3=\frac{x}{3}$

$\Rightarrow \frac{4-2+9}{3}=\frac{x}{3}$

$\Rightarrow \frac{11}{3}=\frac{x}{3}$

$\Rightarrow x=\frac{11 \times 3}{3}=11$

∴x=11

Prove the following (4 to 11) identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

Question 4

(i) $\frac{\cos A}{1-\sin A}+\frac{\cos A}{1+\sin A}=2 \sec A$

(ii) $\frac{\cos A}{\operatorname{cosec} A+1}+\frac{\cos A}{\operatorname{cosec} A-1}=2 \tan A$

Sol :

(i) $\frac{\cos A}{1-\sin A}+\frac{\cos A}{1+\sin A}=2 \sec A$

L.H.S$=\frac{\cos A}{1-\sin A}+\frac{\cos A}{1+\sin A}$

$=\cos \mathrm{A}\left[\frac{1}{1-\sin \mathrm{A}}+\frac{1}{1+\sin \mathrm{A}}\right]$

$=\cos \mathrm{A}\left[\frac{1+\sin \mathrm{A}+1-\sin \mathrm{A}}{(1-\sin \mathrm{A})(1+\sin \mathrm{A})}\right]$

$=\cos \mathrm{A}\left[\frac{2}{1-\sin ^{2} \mathrm{~A}}\right]=\frac{2 \cos \mathrm{A}}{\cos ^{2} \mathrm{~A}}$

$=\frac{2}{\cos A}=2 \sec A$=R.H.S

(ii) $\frac{\cos A}{\operatorname{cosec} A+1}+\frac{\cos A}{\operatorname{cosec} A-1}=2 \tan A$

L.H.S$=\frac{\cos A}{\operatorname{cosec} A+1}+\frac{\cos A}{\operatorname{cosec} A-1}$

$=\cos A\left[\frac{1}{\operatorname{cosec} A+1}+\frac{1}{\operatorname{cosec} A-1}\right]$

$=\cos A\left[\frac{\operatorname{cosec} A-1+\operatorname{cosec} A+1}{(\operatorname{cosec} A+1)(\operatorname{cosec} A-1)}\right]$

$=\frac{\cos A[2 \operatorname{cosec} A]}{\operatorname{cosec}^{2} A-1}=\frac{2 \cos A}{\sin A\left(\cot ^{2} A\right)}$

$=\frac{2 \cot A}{\cot ^{2} A}=\frac{2}{\cot A}$

=2 tan A= R.H.S

Question 5

(i) $\frac{(\cos \theta-\sin \theta)(1+\tan \theta)}{2 \cos ^{2} \theta-1}=\sec \theta$

(ii) (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) = 1.

Sol :

(i) $\frac{(\cos \theta-\sin \theta)(1+\tan \theta)}{2 \cos ^{2} \theta-1}=\sec \theta$

L.H.S $=\frac{(\cos \theta-\sin \theta)(1+\tan \theta)}{2 \cos ^{2} \theta-1}$

$=\frac{(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)}{\cos \theta\left(2 \cos ^{2} \theta-1\right)}$

$=\frac{\cos ^{2} \theta-\sin ^{2} \theta}{\cos \theta\left(2 \cos ^{2} \theta-1\right)}$

$=\frac{\cos ^{2} \theta-\left(1-\cos ^{2} \theta\right)}{\cos \theta\left(2 \cos ^{2} \theta-1\right)}$

$=\frac{\cos ^{2} \theta-1+\cos ^{2} \theta}{\cos \theta\left(2 \cos ^{2} \theta-1\right)}$

$=\frac{2 \cos ^{2} \theta-1}{\cos \theta\left(2 \cos ^{2} \theta-1\right)}$

$=\frac{1}{\cos \theta}=\sec \theta$=R.H.S

(ii) $(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)(\tan \theta+\cot \theta)=1$

$(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)(\tan \theta+\cot \theta)=1$

L.H.S. $=(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)(\tan \theta+\cot \theta)$

$=\left(\frac{1}{\sin \theta}-\sin \theta\right)\left(\frac{1}{\cos \theta}-\cos \theta\right)\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)$

$=\left(\frac{1-\sin ^{2} \theta}{\sin \theta}\right)\left(\frac{1-\cos ^{2} \theta}{\cos \theta}\right)\left(\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}\right)$

$=\frac{\left(\cos ^{2} \theta \times \sin ^{2} \theta\right)}{\sin \theta \cos \theta} \times \frac{1}{\sin \theta \cos \theta}$

$=\frac{\sin ^{2} \cos ^{2} \theta}{\sin ^{2} \theta \cos ^{2} \theta}=1$=R.H.S

Question 6

(ii) $\sin ^{2} \theta+\cos ^{4} \theta=\cos ^{2} \theta+\sin ^{4} \theta$

(ii) $\frac{\cot \theta}{\operatorname{cosec} \theta+1}+\frac{\operatorname{cosec} \theta+1}{\cot \theta}=2 \sec \theta$

Sol :

L.H.S$=\sin ^{2} \theta+\cos ^{4} \theta$

$=\left(1-\cos ^{2} \theta+\cos ^{4} \theta\right.$

$=1-\cos ^{2} \theta+\cos ^{4} \theta$

$=1-\cos ^{2} \theta\left(1-\cos ^{2} \theta\right)$

R.H.S$=\cos ^{2} \theta+\sin ^{4} \theta$

$=1-\sin ^{2} \theta+\sin ^{4} \theta$

$=1-\sin ^{2} \theta\left(1-\sin ^{2} \theta\right)$

$=1-\sin ^{2} \theta \cos ^{2} \theta$

∴L.H.S=R.H.S

(ii) $\frac{\cot \theta}{\operatorname{cosec} \theta+1}+\frac{\operatorname{cosec} \theta+1}{\cot \theta}=2 \sec \theta$

L.H.S$=\frac{\cot \theta}{\operatorname{cosec} \theta+1}+\frac{\operatorname{cosec} \theta+1}{\cot \theta}$

$=\frac{\frac{\cos \theta}{\sin \theta}}{\frac{1}{\sin \theta}+1}+\frac{\frac{1}{\sin \theta}+1}{\frac{\cos \theta}{\sin \theta}}$

$=\frac{\frac{\cos \theta}{\sin \theta}}{\frac{1+\sin \theta}{\sin \theta}}+\frac{\frac{1+\sin \theta}{\sin \theta}}{\frac{\cos \theta}{\sin \theta}}$

$=\frac{\cos \theta}{\sin \theta} \times \frac{\sin \theta}{1+\sin \theta}+\frac{1+\sin \theta}{\sin \theta} \times \frac{\sin \theta}{\cos \theta}$

$=\frac{\cos \theta}{1+\sin \theta}+\frac{1+\sin \theta}{\cos \theta}$

$=\frac{\cos ^{2} \theta+1+\sin ^{2} \theta+2 \sin \theta}{(1+\sin \theta) \cos \theta}$

$=\frac{1+1+2 \sin \theta}{(1+\sin \theta) \cos \theta}=\frac{2+2 \sin \theta}{(1+\sin \theta) \cos \theta}$

$=\frac{2(1+\sin \theta)}{\cos \theta(1+\sin \theta)}=\frac{2}{\cos \theta}$

=2 sec θ=R.H.S

Question 7

(i) $\sec ^{4} A\left(1-\sin ^{4} A\right)-2 \tan ^{2} A=1$

(ii) $\frac{1}{\sin A+\cos A+1}+\frac{1}{\sin A+\cos A-1}=\sec A+\operatorname{cosec} A$

Sol :

(i) $\sec ^{4} \mathrm{~A}\left(1-\sin ^{4} \mathrm{~A}\right)-2 \tan ^{2} \mathrm{~A}=1$

L.H.S$=\sec ^{4} A\left(1-\sin ^{4} A\right)-2 \tan ^{2} A$

$=\frac{1}{\cos ^{4} A}\left(1+\sin ^{2} A\right)\left(1-\sin ^{2} A\right)-2 \tan ^{2} A$

$\left[\because a^{2}-b^{2}=(a+b)(a-b)\right]$

$=\frac{\left(1+\sin ^{2} A\right) \cos ^{2} A}{\cos ^{4} A}-2 \frac{\sin ^{2} A}{\cos ^{2} A}$

$=\frac{1+\sin ^{2} A}{\cos ^{2} A}-\frac{2 \sin ^{2} A}{\cos ^{2} A}$

$=\frac{1+\sin ^{2} A-2 \sin ^{2} A}{\cos ^{2} A}$

$=\frac{1-\sin ^{2} A}{\cos ^{2} A}$

$=\frac{\cos ^{2} A}{\cos ^{2} A}=1$

$\left(\because 1-\sin ^{2} A=\cos ^{2} A\right)$

=R.H.S

(ii) $\frac{1}{\sin A+\cos A+1}+\frac{1}{\sin A+\cos A-1}$

=sec A+cosec A

L.H.S$=\frac{1}{\sin A+\cos A+1}+\frac{1}{\sin A+\cos A-1}$

$=\frac{\sin A+\cos A-1+\sin A+\cos A+1}{(\sin A+\cos A+1)(\sin A+\cos A-1)}$

$=\frac{2(\sin A+\cos A)}{(\sin A+\cos A)^{2}-(1)^{2}}$

$=\frac{2(\sin A+\cos A)}{\sin ^{2} A+\cos ^{2} A+2 \sin A \cos A-1}$

$=\frac{2(\sin A+\cos A)}{1+2 \sin A \cos A-1}$

$=\frac{2(\sin A+\cos A)}{2 \sin A \cos A}$

$=\frac{\sin A+\cos A}{\sin A \cos A}$

$=\frac{\sin A}{\sin A \cos A}+\frac{\cos A}{\sin A \cos A}$

$=\frac{1}{\cos A}+\frac{1}{\sin A}$

=sec A+cosec A=R.H.S

Question 8

(i) $\frac{\sin ^{3} \theta+\cos ^{3} \theta}{\sin \theta \cos \theta}+\sin \theta \cos \theta=1$

(ii) $(\sec A-\tan A)^{2}(1+\sin A)=1-\sin A$

Sol :

(i) $\frac{\sin ^{3} \theta+\cos ^{3} \theta}{\sin \theta \cos \theta}+\sin \theta \cos \theta=1$

L.H.S$=\frac{\sin ^{3} \theta+\cos ^{3} \theta}{\sin \theta \cos \theta}+\sin \theta \cos \theta$

$=\frac{(\sin \theta+\cos \theta)\left(\sin ^{2} \theta-\sin \theta \cos \theta+\cos ^{2} \theta\right)}{(\sin \theta+\cos \theta)}$$+\sin \theta \cos \theta$

$=\sin ^{2} \theta+\cos ^{2} \theta-\sin \theta \cos \theta+\sin \theta \cos \theta$

=1=R.H.S

(ii) $(\sec A-\tan A)^{2}(1+\sin A)^{\doteq}=1-\sin A$

L.H.S$=\left(\frac{1}{\cos A}-\frac{\sin A}{\cos A}\right)^{2}(1+\sin A)$

$=\left(\frac{1-\sin \mathrm{A}}{\cos \mathrm{A}}\right)^{2}(1+\sin \mathrm{A})$

$=\frac{(1-\sin A)^{2}}{\cos ^{2} A}(1+\sin A)$

$=\frac{(1-\sin \mathrm{A})^{2}(1+\sin \mathrm{A})}{1-\sin ^{2} \mathrm{~A}}$

$=\frac{(1-\sin \mathrm{A})^{2}(1+\sin \mathrm{A})}{(1-\sin \mathrm{A})(1+\sin \mathrm{A})}$

=1-sin A=R.H.S

Question 9

(i) $\frac{\cos A}{1-\tan A}-\frac{\sin ^{2} A}{\cos A-\sin A}=\sin A+\cos A$

(ii) (sec A – cosec A) (1 + tan A + cot A) = tan A sec A – cot A cosec A

(iii) $\frac{\tan ^{2} \theta}{\tan ^{2} \theta-1}-\frac{\operatorname{cosec}^{2} \theta}{\sec ^{2} \theta-\operatorname{cosec}^{2} \theta}=\frac{1}{\sin ^{2} \theta-\cos ^{2} \theta}$

Sol :

(i) $\frac{\cos A}{1-\tan A}-\frac{\sin ^{2} A}{\cos A-\sin A}=\sin A+\cos A$

L.H.S$=\frac{\cos A}{1-\tan A}-\frac{\sin ^{2} A}{\cos A-\sin A}$

$=\frac{\cos A}{1-\frac{\sin A}{\cos A}}-\frac{\sin ^{2} A}{\cos A-\sin A}$

$=\frac{\cos A \times \cos A}{\cos A-\sin A}-\frac{\sin ^{2} A}{\cos A-\sin A}$

$=\frac{\cos ^{2} A}{\cos A-\sin A}-\frac{\sin ^{2} A}{\cos A-\sin A}$

$=\frac{\cos ^{2} A-\sin ^{2} A}{\cos A-\sin A}$

$=\frac{(\cos A+\sin A)(\cos A-\sin A)}{(\cos A-\sin A)}$

=cos A+sin A

=sin A+cos A=R.H.S

(ii) $(\sec A-\operatorname{cosec} A)(1+\tan A+\cot A)=\tan A\sec A-\cot A \operatorname{cosec} A$

L.H.S=(secA-cosecA)(1+tanA+cotA)

$=\left(\frac{1}{\cos A}-\frac{1}{\sin A}\right)\left(1+\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right)$

$=\frac{\sin A-\cos A}{\sin A \cos A}\times \frac{\sin A \cos A+\sin ^{2} A+\cos ^{2} A}{\sin A \cos A}$

$=\frac{\sin A-\cos A}{\sin A \cos A} \times \frac{\sin A \cos A+1}{\sin A \cos A}$

$=\frac{(\sin A-\cos A)(\sin A \cos A+1)}{\sin ^{2} A \cos ^{2} A}$

$=\frac{\sin A}{\cos A} \cdot \frac{1}{\cos A}-\frac{\cos A}{\sin A} \cdot \frac{1}{\sin A}$

$=\frac{\sin A}{\cos ^{2} A}-\frac{\cos A}{\sin ^{2} A}$

$=\frac{\sin ^{3} A-\cos ^{3} A}{\sin ^{2} A \cos ^{2} A}$

$=\frac{(\sin A-\cos A)\left(\sin ^{2} A+\cos ^{2} A+\sin A \cos A\right)}{\sin ^{2} A \cos ^{2} A}$

$=\frac{(\sin \mathrm{A}-\cos A)(1+\sin \mathrm{A} \cos \mathrm{A})}{\sin ^{2} \mathrm{~A} \cos ^{2} \mathrm{~A}}$

$=\frac{(\sin \mathrm{A}-\cos \mathrm{A})(\sin \mathrm{A} \cos \mathrm{A}+1)}{\sin ^{2} \mathrm{~A} \cos ^{2} \mathrm{~A}}$

∴L.H.S=R.H.S

(iii) $\frac{\tan ^{2} \theta}{\tan ^{2} \theta-1}+\frac{\operatorname{cosec}^{2} \theta}{\sec ^{2} \theta-\operatorname{cosec}^{2} \theta}=\frac{1}{\sin ^{2} \theta-\cos ^{2} \theta}$

L.H.S$=\frac{\tan ^{2} \theta}{\tan ^{2} \theta-1}+\frac{\operatorname{cosec}^{2} \theta}{\sec ^{2} \theta-\operatorname{cosec}^{2} \theta}$

$=\frac{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}-1}+\frac{\frac{1}{\sin ^{2} \theta}}{\frac{1}{\cos ^{2} \theta}-\frac{1}{\sin ^{2} \theta}}$

$=\frac{\sin ^{2} \theta}{\cos ^{2} \theta} \times \frac{\cos ^{2} \theta}{\sin ^{2} \theta-\cos ^{2} \theta}+\frac{\frac{1}{\sin ^{2} \theta}}{\frac{\sin ^{2} \theta-\cos ^{2} \theta}{\sin ^{2} \theta \cos ^{2} \theta}}$

$=\frac{\sin ^{2} \theta}{\sin ^{2} \theta-\cos ^{2} \theta}+\frac{\sin ^{2} \theta \cos ^{2} \theta}{\sin ^{2} \theta\left(\sin ^{2} \theta-\cos ^{2} \theta\right)}$

$=\frac{\sin ^{2} \theta}{\sin ^{2} \theta-\cos ^{2} \theta}+\frac{\cos ^{2} \theta}{\sin ^{2} \theta-\cos ^{2} \theta}$

$=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin ^{2} \theta-\cos ^{2} \theta}=\frac{1}{\sin ^{2} \theta-\cos ^{2} \theta}$

=R.H.S

Question 10

$\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{2}{\sin ^{2} A-\cos ^{2} A}=\frac{2}{1-2 \cos ^{2} A}=\frac{2 \sec ^{2} A}{\tan ^{2} A-1}$

Sol :

$\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{2}{\sin ^{2} A-\cos ^{2} A}=\frac{2}{1-2 \cos ^{2} A}=\frac{2 \sec ^{2} A}{\tan ^{2} A-1}$

L.H.S$=\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}$

$=\frac{(\sin \mathrm{A}+\cos \mathrm{A})^{2}+(\sin \mathrm{A}-\cos \mathrm{A})^{2}}{\sin ^{2} \mathrm{~A}-\cos ^{2} \mathrm{~A}}$

$=\frac{2\left(\sin ^{2} A+\cos ^{2} A\right)}{\sin ^{2} A-\cos ^{2} A}$

$=\frac{2 \times 1}{\sin ^{2} A-\cos ^{2} A}=\frac{2}{\sin ^{2} A-\cos ^{2} A}$

=R.H.S

Now, we have

$=\frac{2}{\sin ^{2} A-\cos ^{2} A}$

$=\frac{2}{\sin ^{2} A-\left(1-\sin ^{2} A\right)}$

$=\frac{2}{\sin ^{2} A-1+\sin ^{2} A}$

$=\frac{2}{\sin ^{2} A-\cos ^{2} A}$

$=\frac{\frac{2}{\cos ^{2} A}}{\frac{\sin ^{2} A}{\cos ^{2} A}-\frac{\cos ^{2} A}{\cos ^{2} A}}$

$=\frac{2 \sec ^{2} A}{\tan ^{2} A-1}$

Question 11

$2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)+1=\theta$

Sol :

$2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)+1=\theta$

L.H.S=$=2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)+1$

$=2\left[\left(\sin ^{2} \theta\right)^{3}+\left(\cos ^{2} \theta\right)^{3}\right]-3\left[\sin ^{4} \theta+\cos ^{4} \theta\right]+1$

$=2\left[\left(\sin ^{2} \theta+\cos ^{2} \theta\right)\right.\left.\left(\sin ^{4} \theta+\cos ^{4} \theta-\sin ^{2} \theta \cos ^{2} \theta\right)\right]-3\left[\sin ^{4} \theta+\cos ^{4} \theta\right]+1$

$=2\left(\sin ^{4} \theta+\cos ^{4} \theta-\sin ^{2} \theta \cos ^{2} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)+1$

$=-2 \sin ^{2} \theta \cos ^{2} \theta-\sin ^{4} \theta-\cos ^{4} \theta+1$

$=1-\left[\sin ^{4} \theta+\cos ^{4} \theta+2 \sin ^{2} \theta \cos ^{2} \theta\right]$

$=1-\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}=1-1=0$

=R.H.S

Question 12

If cot θ + cos θ = m, cot θ – cos θ = n, then prove that $\left(m^{2}-n^{2}\right)^{2}$= 16 run.

Sol :

cot θ + cos θ = m…..(i)

cot θ – cos θ = n……(ii)

Adding (i)&(ii) we get

$2 \cot \theta=m+n$

$\Rightarrow \cot \theta=\frac{m+n}{2}$

$\therefore \tan \theta=\frac{2}{m+n}$...(i)

Subtracting (ii) from (i)

$2 \cos \theta=m-n \Rightarrow \cos \theta=\frac{m-n}{2}$

$\therefore \quad \sec \theta=\frac{2}{m-n}$...(ii)

Squaring and subtracting

$\sec ^{2} \theta-\tan ^{2} \theta=\left(\frac{2}{m-n}\right)^{2}-\left(\frac{2}{m+n}\right)^{2}$

$1=\frac{4}{(m-n)^{2}}-\frac{4}{(m+n)^{2}}$

$\Rightarrow 4\left[\frac{1}{(m-n)^{2}}-\frac{1}{(m+n)^{2}}\right]=1$

$\Rightarrow 4\left[\frac{(m+n)^{2}-(m-n)^{2}}{(m+n)^{2}(m-n)^{2}}\right]=1$

$\Rightarrow \frac{4(4 m n)}{\left(m^{2}-n^{2}\right)^{2}}=1$

$\Rightarrow \frac{16 m n}{\left(m^{2}-n^{2}\right)^{2}}=1$

$\Rightarrow \quad\left(m^{2}-n^{2}\right)^{2}=16 \mathrm{mn}$

Question 13

If sec θ + tan θ = p, prove that sin θ = $\frac{p^{2}-1}{p^{2}+1}$

Sol :

sec θ + tan θ = p,

prove that sin θ $=\frac{p^{2}-1}{p^{2}+1}$

$\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}=p$

$\frac{1+\sin \theta}{\cos \theta}=p$

Squaring ,

$\frac{(1+\sin \theta)^{2}}{\cos ^{2} \theta}=p^{2}$

$\Rightarrow \frac{(1+\sin \theta)^{2}}{1-\sin ^{2} \theta}=p^{2}$

$\Rightarrow \frac{(1+\sin \theta)^{2}}{(1-\sin \theta)(1+\sin \theta)}=p^{2}$

$\Rightarrow \frac{1+\sin \theta}{1-\sin \theta}=\frac{p^{2}}{1}$

Applying componendo and dividendo

$\frac{1+\sin \theta+1-\sin \theta}{1+\sin \theta-1+\sin \theta}=\frac{p^{2}+1}{p^{2}-1}$

$\frac{1+\sin \theta+1-\sin \theta}{1+\sin \theta-1+\sin \theta}$

$=\frac{p^{2}+1}{p^{2}-1}$

$\therefore \sin \theta=\frac{p^{2}-1}{p^{2}+1}$

Question 14

If tan A = n tan B and sin A = m sin B, prove that $\cos ^{2} A=\frac{m^{2}-1}{n^{2}-1}$

Sol :

$m=\frac{\sin A}{\sin B}$

$n=\frac{\tan A}{\tan B}$

$\frac{1}{\sin \mathrm{B}}=\frac{m}{\sin \mathrm{A}}$

$\Rightarrow \operatorname{cosec} \mathrm{B}=\frac{m}{\sin \mathrm{A}}$

$\frac{1}{\tan \mathrm{B}}=\frac{n}{\tan \mathrm{A}}$

$\Rightarrow \cot \mathrm{B}=\frac{n}{\tan \mathrm{A}}$

Now, $\operatorname{cosec}^{2} B-\cot ^{2} B=1$

$\Rightarrow \frac{m^{2}}{\sin ^{2} A}-\frac{n^{2}}{\tan ^{2} A}=1$

$\Rightarrow \frac{m^{2}}{\sin ^{2} A}-\frac{n^{2} \cos ^{2} A}{\sin ^{2} A}=1$

$\Rightarrow m^{2}-n^{2} \cos ^{2} A=\sin ^{2} A$

$\Rightarrow m^{2}-n^{2} \cos ^{2} A=1-\cos ^{2} A$

$\Rightarrow m^{2}-1=\cos ^{2} A+n^{2} \cos ^{2} A$

$\Rightarrow m^{2}-1=\left(n^{2}-1\right) \cos ^{2} A$

$\Rightarrow \cos ^{2} A=\frac{m^{2}-1}{n^{2}-1}$

Question 15

If sec A  $=x+\frac{1}{4 x}$, , then prove that sec A + tan A = 2x or $\frac{1}{2 x}$

Sol :

$\sec A=x+\frac{1}{4 x}$

To prove that sec A + tan A = 2x or $\frac{1}{2 x}$

$\tan A=\sqrt[\pm]{\sec ^{2} A-1}$

$=\sqrt[\pm]{\left(x+\frac{1}{4 x}\right)^{2}-1}$

$=\sqrt[\pm]{x^{2}+\frac{1}{16 x^{2}}+\frac{1}{2}-1}$

$=\sqrt[\pm]{x^{2}+\frac{1}{16 x^{2}}-\frac{1}{2}}$

$=\pm\left(x-\frac{1}{4 x}\right)$

$\therefore \sec A+\tan A=x+\frac{1}{4 x}+x-\frac{1}{4 x}=2 x$

or $x+\frac{1}{4 x}-x+\frac{1}{4 x}=\frac{1}{2 x}$

Hence proved

Question 16

When 0° < θ < 90°, solve the following equations:

(i) $2 \cos ^{2} \theta+\sin \theta-2=0$

(ii) $3 \cos \theta=2 \sin ^{2} \theta$

(iii) $\sec ^{2} \theta-2 \tan \theta=0$

(iv) $\tan ^{2} \theta=3(\sec \theta-1)$

Sol :

0° < θ < 90°

(i) $2 \cos ^{2} \theta+\sin \theta-2=0$

$2\left(1-\sin ^{2} \theta\right)+\sin \theta-2=0$

$2-2 \sin ^{2} \theta+\sin \theta-2=0$

$-2 \sin ^{2} \theta+\sin \theta=0$

$\Rightarrow \sin \theta(-2 \sin \theta+1)=0$

Either $\sin \theta=0,$ then $\theta=0$

or $-2 \sin \theta+1=0,$ then $\sin \theta=\frac{1}{2}$

$\therefore \theta=30^{\circ}$

Hence $\theta=0^{\circ}$ or $30^{\circ}$

(ii) $3 \cos \theta=2 \sin ^{2} \theta$

$\Rightarrow 3 \cos \theta=2\left(1-\cos ^{2} \theta\right)$

$\Rightarrow 3 \cos \theta=2-2 \cos ^{2} \theta$

$\Rightarrow 2 \cos ^{2} \theta+3 \cos \theta-2=0$

$\Rightarrow 2 \cos ^{2} \theta+4 \cos \theta-\cos \theta-2=0$

$\Rightarrow 2 \cos \theta(\cos \theta+2)-1(\cos \theta+2)=0$

$(\cos \theta+2)(2 \cos \theta-1)=0$

Either $\cos \theta+2=0 \Rightarrow \cos \theta=-2$ which is not possible being negative.

or $2 \cos \theta-1=0, \cos \theta=\frac{1}{2}=\cos 60^{\circ}$

$\therefore \theta=60^{\circ}$

(iii) $\sec ^{2} \theta-2 \tan \theta=0$

$1+\tan ^{2} \theta-2 \tan \theta=0$

$\Rightarrow \tan ^{2} \theta-2 \tan \theta+1=0$

$(\tan \theta-1)^{2}=0 \Rightarrow \tan \theta-1=0$

$\Rightarrow \tan \theta=1=\tan 45^{\circ}$

$\therefore \theta=45^{\circ}$

(iv) $\tan ^{2} \theta=3(\sec \theta-1)$

$\left(\sec ^{2} \theta-1\right)=3(\sec \theta-1)$

$\Rightarrow \sec ^{2} \theta-1=3 \sec \theta-3$

$\Rightarrow \sec ^{2} \theta-1-3 \sec \theta+3=0$

$\Rightarrow \sec ^{2} \theta-3 \sec \theta+2=0$

$\Rightarrow \sec ^{2} \theta-\sec \theta-2 \sec \theta+2=0$

$\Rightarrow \sec \theta(\sec \theta-1)-2(\sec \theta-1)=0$

$\Rightarrow(\sec \theta-1)(\sec \theta-2)=0$

Either $\sec \theta-1=0, \sec \theta=1=\sec 0^{\circ}$

$\therefore \theta=0^{\circ}$ not possible $\left(\because \theta>0^{\circ}\right)$

or $\sec \theta-2=0 \Rightarrow \sec \theta=2$

$\therefore \sec \theta=2=\sec 60^{\circ}$

$\therefore \theta=60^{\circ}$