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प्रश्न
Prove that `(1 + sin B)/(cos B) + (cos B)/(1 + sin B) = 2 sec B`.
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उत्तर
L.H.S = `(1 + sin B)/(cos B) + (cos B)/(1 + sin B)`
= `((1 + sin B)^2 + cos^2B)/(cos B(1 + sin B))`
= `(1 + 2 sin B + sin^2B + cos^2B)/(cos B(1 + sin B))` ...[∵ (a + b)2 = a2 + 2ab + b2]
= `(1 + 2 sin B + 1)/(cos B(1 + sin B))` ...[∵ sin2B + cos2B = 1]
= `(2 + 2 sin B)/(cos B(1 + sin B))`
= `(2(1 + sin B))/(cos B(1 + sin B))`
= `2/(cos B)`
= 2 sec B
= R.H.S.
∴ `(1 + sin B)/(cos B) + (cos B)/(1 + sin B) = 2 sec B`
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∴ 1 + tan2 θ = `(25/7)^square`
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= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
