Question

1+tan^2Q / 1+ cot^2 = ( 1-tanQ/1-cotQ )^2​

Answer 1

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

Let  $\theta$  be A

$\boxed{\bold{\frac{1+tan^2A}{1+cot^2A} = (\frac{1-tanaA}{1-cotA})^2=tan^2A}}$

At first,we have to prove that   ${\bold{\frac{1+tan^2A}{1+cot^2A} =tan^2A}$

(1)  ${\bold{\frac{1+tan^2A}{1+cot^2A} =tan^2A}$

Taking L.H.S,

${\bold{\frac{1+tan^2A}{1+cot^2A}}$

[ 1+tan² A = sec² A ]

[ 1+cot² A = cosec² A ]

$\implies{\bold{\frac{1+tan^2A}{1+cot^2A} =\frac{sec^2A}{cosec^2A}}$

[ sec²A = 1/cos²A ]

[ cosec²A = 1 / sin²A ]

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

[sin² A /cos² A= tan² A]

$\boxed{\bold{\therefore\frac{1+tan^2A}{1+cot^2A} =tan^2A}}$

Now, we have to prove that $\bold{(\frac{1-tanA}{1-cotA})^2=tan^2A}}$

(ii) $\bold{(\frac{1-tanA}{1-cotA})^2=tan^2A}}$

Taking L.H.S,

$\bold{(\frac{1-tanA}{1-cotA})^2}$

$\implies\bold{\frac{1-\frac{sinA}{cosA}}{(1-\frac{cosA}{sinA})^2}}$

$\implies\bold{\frac{\frac{cosA-sinA}{cosA}}{(\frac{sinA-cosA}{sinA})^2}}$

$\implies\bold{\frac{\frac{1}{cosA}}{(\frac{1}{sinA})^2}}$

$\implies{(\frac{sinA}{cosA})^2$

$\implies(tanA)^2$

$\boxed{\bold{\therefore(\frac{1-tanA}{1-cotA})^2=tan^2A}}}$

Hence Proved !!