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प्रश्न
Prove that `tan^3 θ/( 1 + tan^2 θ) + cot^3 θ/(1 + cot^2 θ) = sec θ. cosec θ - 2 sin θ cos θ.`
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उत्तर
LHS = `tan^3 θ/(1 + tan^2 θ) + cot^3 θ/(1 + cot^2 θ)`
= `tan^3 θ/sec^2 θ + cot^3 θ/(cosec^2 θ)`
= 1 + tan2θ = sec2θ; 1 + cot2θ = cosec2θ
= `sin^3 θ/cos^3 θ xx cos^2 θ + cos^3 θ/sin^3 θ xx sin^2 θ`
= `sin^3 θ/cos θ + cos^3 θ/sin θ`
= `(sin^4 θ + cos^4 θ)/(cos θ.sin θ)`
= `((sin^2θ)^2 + (cos^2θ)^2)/(sin θ.cos θ)`
= `((sin^2 θ + cos^2 θ)^2 - 2 sin^2 θ. cos^2 θ)/(sin θ.cos θ)` ...[a2 + b2 = (a + b)2 − 2ab]
= `((1)^2 - 2sin^2θ. cos^2θ)/(sin θ.cos θ)`
= `(1 - 2sin^2θ. cos^2θ)/(sinθ.cosθ)`
= `1/(sinθ.cosθ) - (2sin^2θ. cos^2θ)/(sinθ.cosθ)`
= secθ. cosecθ − 2 sinθ cosθ
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
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`(1 + sec theta)/sec theta = (sin^2 theta)/(1 - cos theta)`
Prove the following trigonometric identities.
`(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`
Prove the following identities:
`(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(2sin^2A - 1)`
`costheta/((1-tan theta))+sin^2theta/((cos theta-sintheta))=(cos theta+ sin theta)`
If cosec θ = 2x and \[5\left( x^2 - \frac{1}{x^2} \right)\] \[2\left( x^2 - \frac{1}{x^2} \right)\]
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The value of the expression \[\sin {80}^° - \cos {80}^°\]
Prove the following identity :
`(cot^2θ(secθ - 1))/((1 + sinθ)) = sec^2θ((1-sinθ)/(1 + secθ))`
If `(cos alpha)/(cos beta)` = m and `(cos alpha)/(sin beta)` = n, then prove that (m2 + n2) cos2 β = n2
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`
