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प्रश्न
If x = asecθ + btanθ and y = atanθ + bsecθ , prove that `x^2 - y^2 = a^2 - b^2`
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उत्तर
`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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संबंधित प्रश्न
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`sin^2Acos^2B - cos^2Asin^2B = sin^2A - sin^2B`
If sec θ + tan θ = m, show that `(m^2 - 1)/(m^2 + 1) = sin theta`
Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.
Which is not correct formula?
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
