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प्रश्न
Prove the following Identities :
`(cosecA)/(cotA+tanA)=cosA`
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उत्तर
`(cosecA)/(cotA+tanA)=cosA`
= LHS
= `(cosecA)/(cotA+tanA)`
= `(cosecA)/(cosA/sinA+sinA/cosA)`
=`((cosecA)/(cos^2A+sin^2A))/(sinA.cosA)`
= `(1/sinA)/(1/(sinA.cosA))`
= `(sinA.cosA)/sinA`
= cosA
= RHS
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संबंधित प्रश्न
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`(1+ tan^2 theta)/(1+ tan^2 theta)= (cos^2 theta - sin^2 theta)`
Prove the following identity :
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If A + B = 90°, show that `(sin B + cos A)/sin A = 2tan B + tan A.`
Prove the following:
`1 + (cot^2 alpha)/(1 + "cosec" alpha)` = cosec α
Prove the following that:
`tan^3θ/(1 + tan^2θ) + cot^3θ/(1 + cot^2θ)` = secθ cosecθ – 2 sinθ cosθ
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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`
