Question

prove it . Trigonometry​

Answer 1

well your question is a bit incorrect. It shouldn't be 'whole square' at theR.H.S..

I guess it's a typing mistake.

Answer 2

Answer :

Given :

To prove that,

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

We know that,

$tan A = \frac{sinA}{cosA}$  ..(i)

$cot A = \frac{cosA}{sinA}$  ..(ii)

$sin^2A+cos^2A = 1$           ..(iii)

Step-by-step explanation:

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

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

= $\frac{1+(\frac{sinA}{cosA})^2}{1+(\frac{cosA}{sinA})^2}$  [ by (i) & (ii) ]

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

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

= $\frac{\frac{1}{cos^2A}}{\frac{1}{sin^2A}}$ [by (iii) ]

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

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

by multiplying by $(-\frac{(cosA - sinA)}{(sinA - cosA)})^2$,

We get,

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

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

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

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

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

Hence, proved