Question

$\frac{1+{\cot }^2\theta }{1+{\tan }^2\theta }={\left(\frac{1+\cot \theta }{1+\tan \theta }\right)}^2$

• A. True
• B. False

Answer 1

The correct option is A True

Taking LHS

$\frac{1+co{t}^2\theta }{1+ta{n}^2\theta }$

$=\frac{1+\frac{co{s}^2\theta }{si{n}^2\theta }}{1+\frac{si{n}^2\theta }{co{s}^2\theta }}$

$=\frac{co{s}^2\theta (si{n}^2\theta +co{s}^2\theta )}{si{n}^2\theta (si{n}^2\theta +co{s}^2\theta )}$

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

Taking RHS

$(\frac{1+cot\theta }{1+tan\theta }{)}^2$

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

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

$=Co{t}^2\theta -\left(2\right)$

From eqs.(1) and (2)

LHS=RHS

Hence proved.