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प्रश्न
Prove the following trigonometric identities
tan2 A + cot2 A = sec2 A cosec2 A − 2
उत्तर
In the given question, we need to prove tan2 A + cot2 A = sec2 A cosec2 A − 2
Now using `tan theta = sin theta/cos theta` and `cot theta = cos theta/sin theta` in LHS we get
`tan^2 A + cot^2 A = sin^2 A/cos^2 A + cos^2 A/sin^2 A`
`= (sin^4 A + cos^4 A)/(cos^2 A sin^2 A)`
`= ((sin^2 A)^2 + (cos^2 A)^2)/(cos^2 A sin^2 A)`
Further, using the identity `a^2 + b^2 = (a + b)^2 - 2ab` we get
`((sin^2 A)^2 + (cos^2 A)^2)/(cos^2 A sin^2 A) = ((sin^2 A + cos^ A)^2 - 2 sin^2 A cos^2 A)/(sin^2 A cos^2 A)`
`= ((1)^2 - 2sin^2 A cos^2 A)/(sin^2 A cos^2 A)`
`= 1/(sin^2 A cos^2 A) - (2 sin^2 A cos^2 A)/(sin^2 A cos^2 A`
`= cosec^2 A sec^2 A - 2`
Since L.H.S = R.H.S
Hence proved.
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