Question

Prove that : tan2A + cot2A + 2 = sec
2A.cosec
2A

Answer 1

Step-by-step explanation:

$\bold{tan^2A +cot^2A+2=sec^2Acosec^2A}$

Considering L.H.S :-

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

We can write $\bold{tan^2A+cot^2A+2}$ as :-

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

We know :-

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

Transposing 1 to the denominator :-

$\bold{\cfrac{1}{cos^2A}+\cfrac{1}{cosec^2A} }$

Taking L.C.M :-

$\bold{\dfrac{sin^2A+cos^2A}{cos^2Asin^2A} }$

We know :-

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

Substituting we get :-

$\bold{\dfrac{1}{cos^2Asin^2A} }$

We know :-

$\bold{\dfrac{1}{cosA}=secA }$ , $\bold{\dfrac{1}{sinA}=cosecA }$

So we get :-

$\bold{sec^2Acosec^2A}$

∴ L.H.S = R.H.S

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