Question

1+tAn square A/1+cot square A=1-tanAsquare/1-cotA square =tan square A

Answer 1

solution:

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

=$\frac{sec^{2}A}{cosec^{2}A}$

/* we know the Trigonometric identities:

i)1+tan²A = sec²A

ii) 1+cot²A = cosec²A

*/

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

= $\frac{sin^{2}A}{cos^{2}A}$

= $tan^{2}A$ ----(1)

ii) $\left(\frac{1-tanA}{1-cotA}\right)^{2}$

= $\left(\frac{1-tanA}{1-\frac{1}{tanA}}\right)^{2}$

=$\left(\frac{1-tanA}{\frac{tanA-1}{tanA}}\right)^{2}$

= $\left(\frac{1}{\frac{-1}{tanA}}\right)^{2}$

= $(-tanA)^{2}$

=$tan^{2}A$----(2)

Therefore,

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

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

= $tan^{2}A$

••••

Answer 2

To show that

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

Solution:

LHS

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

We know that

$1+tan^2A=sec^2A$

$1+cot^2A=cosec^2A$

Substitute the values

$\frac{sec^2A}{cosec^2A}$

Formula:

$secA=\frac{1}{cosA}$

$cosecA=\frac{1}{sinA}$

Using the formula

$\frac{sin^2A}{cos^2A}=tan^2A$

By using

$tanA=\frac{sinA}{cosA}$

MHS

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

Formula:

$cotA=\frac{cosA}{sinA}$

Using the formula

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

$(\frac{sin^2A}{cos^2A}\times(\frac{cosA-sinA}{sinA-cosA})^2$

$tan^2A(cosA-sinA)^2\times \frac{1}{(-1(cosA-sinA))^2}$

$tan^2A(cosA-sinA)^2\times \frac{1}{(cosA-sinA)^2}$

$tan^2 A$

LHS=MHS=RHS

Hence, proved.