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Question
Prove the following identities:
`(tan"A"+tan"B")/(cot"A"+cot"B")=tan"A"tan"B"`
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Solution
`(tan"A"+tan"B")/(cot"A"+cot"B")=tan"A"tan"B"`
= L.H.S.
=`(tan"A"+tan"B")/(cot"A"+cot"B")`
= `(tan"A"+tan"B")/(1/tanA+1/tanB`
= `(tan"A"+tan"B")/((tan"A"+tan"B")/(tan"A".tan"B"))`
= `((tan"A"+tan"B")(tan"A".tan"B"))/(tan"A"+tan"B")`
= tan A tan B
= R.H.S.
Hence, proved.
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RELATED QUESTIONS
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Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
Prove the following:
`tanA/(1 + sec A) - tanA/(1 - sec A)` = 2cosec A
