Question

Solve:
${\left(1-\tan x\right)}^2+{\left(1-\cot x\right)}^2-{\left(\sec x-\cos ecx\right)}^2$

Answer 1

Consider the given expression.

${(1-\tan x)}^2+{(1-\cot x)}^2+{(\sec x-\cos ecx)}^2$

$=1+{\tan }^{2}x-2\tan x+1+{\cot }^{2}x-2\cot x+{\sec }^{2}x+{\mathrm{cosec}}^{2}x-2secxcosecx$

We know that,

${\sec }^{2}x=1+{\tan }^{2}x$

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

Therefore,

$={\sec }^{2}x-2\tan x+{\mathrm{cosec}}^{2}x-2\cot x+{\sec }^{2}x+{\mathrm{cosec}}^{2}x-2\sec x\text{cosec}x$

$=2{\sec }^{2}x-2(\tan x+\cot x)+2{\mathrm{cosec}}^{2}x-2\sec xcosecx$

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

$=\frac{2({\sin }^{2}x+{\cos }^{2}x)}{si{n}^{2}x{\cos }^{2}x}-2\left(\frac{si{n}^{2}x+{\cos }^{2}x}{sinx\cos x}\right)-\frac{2}{\sin x\cos x}$

$=\frac{2}{si{n}^{2}x{\cos }^{2}x}-\frac{2}{sinx\cos x}-\frac{2}{\sin x\cos x}$

$=\frac{2}{si{n}^{2}x{\cos }^{2}x}-\frac{4}{\sin x\cos x}$

$=\frac{2(1-2\sin x\cos x)}{si{n}^{2}x{\cos }^{2}x}$

$=\frac{2\times 4(1-\sin 2x)}{4si{n}^{2}x{\cos }^{2}x}$

$=\frac{8(1-\sin 2x)}{\sin 2x}$

$=8(\mathrm{cosec}2x-1)$

Hence, the value is $8(\mathrm{cosec}2x-1)$.