# Prove that

Question:

Prove that

(i) $\frac{\cos (2 \pi+x) \operatorname{cosec}(2 \pi+x) \tan (\pi / 2+x)}{\sec (\pi / 2+x) \cos x \cot (\pi+x)}=1$

(ii) $\frac{\operatorname{cosec}\left(90^{\circ}+x\right)+\cot \left(450^{\circ}+x\right)}{\operatorname{cosec}\left(90^{\circ}-x\right)+\tan \left(180^{\circ}-x\right)}+\frac{\tan \left(180^{\circ}+x\right)+\sec \left(180^{\circ}-x\right)}{\tan \left(360^{\circ}+x\right)-\sec (-x)}=2$

(iii) $\frac{\sin \left(180^{\circ}+x\right) \cos \left(90^{\circ}+x\right) \tan \left(270^{\circ}-x\right) \cot \left(360^{\circ}-x\right)}{\sin \left(360^{\circ}-x\right) \cos \left(360^{\circ}+x\right) \operatorname{cosec}(-x) \sin \left(270^{\circ}+x\right)}=1$

(iv) $\left\{1+\cot x-\sec \left(\frac{\pi}{2}+x\right)\right\}\left\{1+\cot x+\sec \left(\frac{\pi}{2}+x\right)\right\}=2 \cot x$

(v) $\frac{\tan \left(90^{\circ}-x\right) \sec \left(180^{\circ}-x\right) \sin (-x)}{\sin \left(180^{\circ}+x\right) \cot \left(360^{\circ}-x\right) \operatorname{cosec}\left(90^{\circ}-x\right)}=1$

Solution:

(i) LHS $=\frac{\cos (2 \pi+x) \cos e c(2 \pi+x) \tan \left(\frac{\pi}{2}+x\right)}{\sec \left(\frac{\pi}{2}+x\right) \cos x \cot (\pi+x)}$

$=\frac{\cos x \operatorname{cosec} x[-\cot x]}{[-\operatorname{cosec} x] \cos x \cot x}$

$=\frac{-\cos x \operatorname{cosec} x \cot x}{-\operatorname{cosec} x \cos x \cot x}$

$=1$

= RHS

Hence, proved.

(ii) $\mathrm{LHS}=\frac{\operatorname{cosec}\left(90^{\circ}+x\right)+\cot \left(450^{\circ}+\mathrm{x}\right)}{\operatorname{cosec}\left(90^{\circ}-x\right)+\tan \left(180^{\circ}-x\right)}+\frac{\tan \left(180^{\circ}+x\right)+\sec \left(180^{\circ}-x\right)}{\tan \left(360^{\circ}+x\right)-\sec (-x)}$

$=\frac{\operatorname{cosec}\left(90^{\circ}+x\right)+\cot \left(450^{\circ}+x\right)}{\operatorname{cosec}\left(90^{\circ}-x\right)+\tan \left(180^{\circ}-x\right)}+\frac{\tan \left(180^{\circ}+x\right)+\sec \left(180^{\circ}-x\right)}{\tan \left(360^{\circ}+x\right)-\sec (-x)}$

$=\frac{\operatorname{cosec}\left(90^{\circ}+x\right)+\cot \left(90^{\circ} \times 5+x\right)}{\operatorname{cosec}\left(90^{\circ}-x\right)+\tan \left(90^{\circ} \times 2-x\right)}+\frac{\tan \left(90^{\circ} \times 2+x\right)+\sec \left(90^{\circ} \times 2-x\right)}{\tan \left(90^{\circ} \times 4+x\right)-\sec (-x)}$

$=\frac{\sec x+\cot \left(90^{\circ} \times 5+x\right)}{\operatorname{cosec}\left(90^{\circ}-x\right)+\tan \left(90^{\circ} \times 2-x\right)}+\frac{\tan \left(90^{\circ} \times 2+x\right)+\sec \left(90^{\circ} \times 2-x\right)}{\tan \left(90^{\circ} \times 4+x\right)-\sec (-x)}$

$=\frac{\sec x-\tan x}{\sec x-\tan x}+\frac{\tan x-\sec x}{\tan x-\sec x}$

$=1+1$

$=2$

= RHS

Hence, proved.

(iii) LHS $=\frac{\sin \left(180^{\circ}+x\right) \cos \left(90^{\circ}+x\right) \tan \left(270^{\circ}-x\right) \cot \left(360^{\circ}-x\right)}{\sin \left(360^{\circ}-x\right) \cos \left(360^{\circ}+x\right) \operatorname{cosec}(-x) \sin \left(270^{\circ}+x\right)}$

$=\frac{\sin \left(90 \times 2^{\circ}+x\right) \cos \left(90^{\circ} \times 1+x\right) \tan \left(90^{\circ} \times 3-x\right) \cot \left(90^{\circ} \times 4-x\right)}{\sin \left(90^{\circ} \times 4-x\right) \cos \left(90^{\circ} \times 4+x\right) \operatorname{cosec}(-x) \sin \left(90^{\circ} \times 3+x\right)}$

$=\frac{-\sin x[-\sin x] \cot x[-\cot x]}{[-\sin x] \cos x[-\cos e c x][-\cos x]}$

$=\frac{\sin ^{2} x \cot ^{2} x}{\sin x \cos e c x \cos x \cos x}$

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

$=\frac{\cos ^{2} x}{\cos ^{2} x}$

$=1$

= RHS

Hence, proved.

(iv) $\mathrm{LHS}=\left\{1+\cot x-\sec \left(\frac{\pi}{2}+x\right)\right\}\left\{1+\cot x+\sec \left(\frac{\pi}{2}+x\right)\right\}$

$=[1+\cot x-\{-\operatorname{cosec} x\}][1+\cot x+\{-\operatorname{cosec} x\}]$

$=[1+\cot x+\operatorname{cosec} x][1+\cot x-\operatorname{cosec} x]$

$=[1+\cot x+\operatorname{cosec} x][1+\cot x-\operatorname{cosec} x]$

$=[\{1+\cot (x)\}+\{\operatorname{cosec} x\}][\{1+\cot x\}-\{\operatorname{cosec} x\}]$

$=\{1+\cot x\}^{2}-\{\operatorname{cosec} x\}^{2}$

$=1+\cot ^{2} x+2 \cot x-\operatorname{cosec}^{2} x$

$=2 \cot x \quad\left[\because 1+\cot ^{2} x=\operatorname{cosec}^{2} x\right]$

= RHS

Hence, proved.

(v) $\mathrm{LHS}=\frac{\tan \left(90^{\circ}-x\right) \sec \left(180^{\circ}-x\right) \sin (-x)}{\sin \left(180^{\circ}+x\right) \cot \left(360^{\circ}-x\right) \operatorname{cosec}\left(90^{\circ}-x\right)}$

$=\frac{\tan \left(90^{\circ} \times 1-x\right) \sec \left(90^{\circ} \times 2-x\right) \sin (-x)}{\sin \left(90^{\circ} \times 2+x\right) \cot \left(90^{\circ} \times 4-x\right) \operatorname{cosec}\left(90^{\circ} \times 1-x\right)}$

$=\frac{\cot x[-\sec x][-\sin x]}{[-\sin x][-\cot x] \sec x}$

$=\frac{\cot x \sec x \sin x}{\sin x \cot x \sec x}$

$=\frac{\frac{\cos x}{\sin x} \times \frac{1}{\cos x} \times \sin x}{\sin x \times \frac{\cos x}{\sin x} \times \frac{1}{\cos x}}$

$=\frac{1}{1}$

$=1$

= RHS

Hence, proved.