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प्रश्न
Prove the following identity :
`(1 + cosA)/(1 - cosA) = (cosecA + cotA)^2`
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उत्तर
`(1 + cosA)/(1 - cosA) = (cosecA + cotA)^2`
= `(1 + cosA)/(1 - cosA).(1 + cosA)/(1 + cosA)`
= `((1 + cosA)^2)/(1 - cos^2A) = (1 + cosA)^2/sin^2A`
= `[(1 + cosA)/sinA]^2 = [1/sinA + cosA/sinA]^2`
= `(cosecA + cotA)^2`
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संबंधित प्रश्न
Express the ratios cos A, tan A and sec A in terms of sin A.
Prove the following identities:
`cotA/(1 - tanA) + tanA/(1 - cotA) = 1 + tanA + cotA`
Prove the following identities:
`sin theta/((cot theta + "cosec" theta)) - sin theta/((cot theta - "cosec" theta)) = 2`
`{1/((sec^2 theta- cos^2 theta))+ 1/((cosec^2 theta - sin^2 theta))} ( sin^2 theta cos^2 theta) = (1- sin^2 theta cos ^2 theta)/(2+ sin^2 theta cos^2 theta)`
Prove that `( 1 + sin θ)/(1 - sin θ) = 1 + 2 tan θ/cos θ + 2 tan^2 θ` .
Prove that `(cos^2θ)/(sinθ) + sin θ = "cosec" θ`.
If cosec A – sin A = p and sec A – cos A = q, then prove that `(p^2q)^(2/3) + (pq^2)^(2/3) = 1`.
Prove the following that:
`tan^3θ/(1 + tan^2θ) + cot^3θ/(1 + cot^2θ)` = secθ cosecθ – 2 sinθ cosθ
Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
