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Question
If x = asecθ + btanθ and y = atanθ + bsecθ , prove that `x^2 - y^2 = a^2 - b^2`
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Solution
`x^2 - y^2 = (asecθ + bTanθ)^2 - (aTanθ + bSecθ)^2`
⇒ `a^2sec^2θ + b^2Tan^2θ + 2abSecθTanθ - (a^2Tan^2θ + b^2Sec^2θ + 2abSecθTanθ)`
⇒ `sec^2θ(a^2 - b^2) + Tan^2θ(b^2 - a^2) = (a^2 - b^2)[Sec^2θ - Tan^2θ]`
⇒ `(a^2 - b^2)` [Since `sec^2θ - Tan^2θ = 1`]
Hence , `x^2 - y^2 = a^2 - b^2`
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RELATED QUESTIONS
Express the ratios cos A, tan A and sec A in terms of sin A.
Prove the following trigonometric identities.
`cos theta/(1 + sin theta) = (1 - sin theta)/cos theta`
Prove the following trigonometric identities.
(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1
Prove the following trigonometric identities.
`(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`
Prove the following identities:
`(1+ sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A) = 2(1 + cot A)`
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
If`( 2 sin theta + 3 cos theta) =2 , " prove that " (3 sin theta - 2 cos theta) = +- 3.`
If `cosec theta = 2x and cot theta = 2/x ," find the value of" 2 ( x^2 - 1/ (x^2))`
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
