Question

Help me for this question ​

Answer 1

Here,

L.H.S = cot^2x/1+cot^2x

= cos^2x/sin^2x / 1+(cos^2x/sin^2x)

= cos^2x/sin^2x / (sin^2x+cos^2x)/sin^2x

= cos^2x/1

=cos^2x

R.H.S proved

Answer 2

See below.

Explanation:

`\boxed{\begin{minipage}{4 cm}\underline{Trigonometric Identities}\\\\$\cot \theta=(\cos \theta)/(\sin \theta)$\\\\\\$\sin^2 \theta +\cos^2 \theta = 1\\\end{minipage}}`

Given expression:

`(\cot^2x)/(1+\cot^2x)`

Rewrite cot²x using the trigonometric identity:

`\implies ((\cos^2x)/(\sin^2x))/(1+(\cos^2x)/(\sin^2x))`

Rewrite the denominator so that is one fraction:

`\implies ((\cos^2x)/(\sin^2x))/((\sin^2x)/(\sin^2x)+(\cos^2x)/(\sin^2x))`

`\implies((\cos^2x)/(\sin^2x))/((\sin^2x+\cos^2x)/(\sin^2x))`

`\textsf{Apply the fraction rule}: \quad ((a)/(c))/((b)/(c))=(a)/(b)`

`\implies (\cos^2x)/(\sin^2x+\cos^2x)`

Use the trigonometric identity sin²x + cos²x = 1 :

`\implies (\cos^2x)/(1)`

Simplify:

`\implies \cos^2x\end{aligned}`

-------------------------------------------------------------------------------------

As one calculation:

`\begin{aligned}(\cot^2x)/(1+\cot^2x)&=((\cos^2x)/(\sin^2x))/(1+(\cos^2x)/(\sin^2x))\\\\&=((\cos^2x)/(\sin^2x))/((\sin^2x)/(\sin^2x)+(\cos^2x)/(\sin^2x))\\\\&=((\cos^2x)/(\sin^2x))/((\sin^2x+\cos^2x)/(\sin^2x))\\\\&=(\cos^2x)/(\sin^2x+\cos^2x)\\\\&=(\cos^2x)/(1)\\\\&=\cos^2x\end{aligned}`