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प्रश्न
If `a cos^3 theta + 3a cos theta sin^2 theta = m, a sin^3 theta + 3 a cos^2 theta sin theta = n`, prove that `(m + n)^(2/3) + (m - n)^(2/3) = 2a^(2/3)`
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उत्तर
`= (a cos^3 theta + 3a cos theta sin^2 theta + a sin^3 theta + 3a cos^2 theta sin theta)^(3/2) + (a cos^3 theta + 3a cos theta sin^2 theta - a sin^3 theta - 3a cos^2 theta sin theta)^(2/3)`
`= a^(1/3) (cos^3 theta + 3 cos theta sin^2 theta + sin^3 theta + 3 cos^2 theta sin theta)^(2/3) + a^(2/3) (cos^3 theta + 3 cos theta sin^2 theta + sin^3 theta - 3 cos^2 theta sin theta)^(2/3)`
`= a^(1/3) [(cos theta + sin theta)^3]^(2/3) + a^(2/3) (cos theta - sin theta)^3]^(2/3)`
`= a^(2/3) [(cos theta + sin theta)^2] + a^(2/3) (cos theta - sin theta)^2`
`= a^(2/3) [cos^2 theta + sin^2 theta - 2sin theta cos theta]`
`= a^(2/3) [cos^2 theta + sin^2 theta + 2 sin theta cos theta] +_ a^(2/3) [cos^2 theta + sin^2 theta - 2 sin theta cos theta]`
`= a^(2/3) [1 + 2 sin theta cos theta] + a^(2/3)[1 - 2 sin theta cos theta]`
`= a^(2/3) [1 + 2 sin theta cos theta + 1 - 2 sin theta cos theta]`
`= a^(1/3) (1 + 1) = 2a^(2/3)`
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संबंधित प्रश्न
Prove the following trigonometric identities.
`1/(1 + sin A) + 1/(1 - sin A) = 2sec^2 A`
Prove the following trigonometric identities.
`((1 + tan^2 theta)cot theta)/(cosec^2 theta) = tan theta`
Prove the following identities:
`(1 + sinA)/cosA + cosA/(1 + sinA) = 2secA`
Prove the following identities:
`1 - cos^2A/(1 + sinA) = sinA`
`1+ (cot^2 theta)/((1+ cosec theta))= cosec theta`
`(1+ tan^2 theta)/(1+ tan^2 theta)= (cos^2 theta - sin^2 theta)`
`sqrt((1-cos theta)/(1+cos theta)) = (cosec theta - cot theta)`
If `secθ = 25/7 ` then find tanθ.
Write 'True' or 'False' and justify your answer in the following:
The value of sin θ + cos θ is always greater than 1.
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`(1 + tan^2A) + (1 + 1/tan^2A) = 1/(sin^2A - sin^4A)`
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`cosec49°cos41° + (tan31°)/(cot59°)`
Choose the correct alternative:
1 + tan2 θ = ?
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If A = 30°, verify that `sin 2A = (2 tan A)/(1 + tan^2 A)`.
Prove that: `cos^2 A + 1/(1 + cot^2 A) = 1`.
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If `cos theta/(1 + sin theta) = 1/"a"`, then prove that `("a"^2 - 1)/("a"^2 + 1)` = sin θ
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`
