Question

Prove the identity :

(1+tan²Ø)/(1+cot²Ø) = [(1-tanØ)/(1-cotØ)]² ​

Answer 1

Answer:

/* Here I am using A instead of Ø */

$RHS=\left(\frac{(1-tanA)}{1-cotA}\right)^{2}\\=\left(\frac{(1-tanA)}{1-\frac{1}{tanA}}\right)^{2}\\=\left(\frac{tanA(1-tanA)}{tanA-1}\right)^{2}\\=\left(\frac{-tanA(tanA-1)}{tanA-1}\right)^{2}\\=(-tanA)^{2}\\=tan^{2}A\\=\frac{sin^{2}A}{cos^{2}A}\\=\frac{sec^{2}A}{cosec^{2}A}$

/* By Trigonometric identities:

i) sec²A = 1+tan²A

ii) cosec²A = 1+cot²A */

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

Therefore,

$\left(\frac{(1-tanA)}{1-cotA}\right)^{2}=\frac{1+tan^{2}A}{1+cot^{2}A}$

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Answer 2

Answer:

RHS=(1−cotA(1−tanA))2=(1−tanA1(1−tanA))2=(tanA−1tanA(1−tanA))2=(tanA−1−tanA(tanA−1))2=(−tanA)2=tan2A=cos2Asin2A=cosec2Asec2A

/* By Trigonometric identities:

i) sec²A = 1+tan²A

ii) cosec²A = 1+cot²A */

\begin{lgathered}=\frac{1+tan^{2}A}{1+cot^{2}A}\\=LHS\end{lgathered}=1+cot2A1+tan2A=LHS

Therefore,

\left(\frac{(1-tanA)}{1-cotA}\right)^{2}=\frac{1+tan^{2}A}{1+cot^{2}A}(1−cotA(1−tanA))2=1+cot2A1+tan2A