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प्रश्न
prove that `1/(1 + cos(90^circ - A)) + 1/(1 - cos(90^circ - A)) = 2cosec^2(90^circ - A)`
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उत्तर
LHS = `1/(1 + cos(90^circ - A)) + 1/(1 - cos(90^circ - A))`
= `1/(1 + sinA) + 1/(1 - sinA)`
= `(1 - sinA + 1 + sinA)/((1 + sinA)(1 - sinA))`
= `2/(1 - sin^2A)`
= `2/cos^2A`
= `2sec^2A = 2cosec^2(90^circ - A)`
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Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
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= R.H.S.
