Math, asked by sumitkumar80927, 13 days ago

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

Answers

Answered by CopyThat
8

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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