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प्रश्न
If m = a sec A + b tan A and n = a tan A + b sec A, then prove that : m2 – n2 = a2 – b2
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उत्तर
Given,
m = a sec A + b tan A and n = a tan A + b sec A
m2 – n2 = (a sec A + b tan A)2 – ( a tan A + b sec A)2
= a2 sec2 A + b2 tan2 A + 2ab sec A tan A – (a2 tan2 A + b2 sec2 A + 2ab sec A tan A)
= sec2 A (a2 – b2) + tan2 A (b2 – a2)
= (a2 – b2) [sec2 A – tan2 A]
= (a2 – b2) [Since sec2 A – tan2 A = 1]
Hence, m2 – n2 = a2 – b2
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संबंधित प्रश्न
Prove the following trigonometric identities.
`cos theta/(1 + sin theta) = (1 - sin theta)/cos theta`
Prove the following identities:
`1/(1 - sinA) + 1/(1 + sinA) = 2sec^2A`
(i)` (1-cos^2 theta )cosec^2theta = 1`
`sin theta / ((1+costheta))+((1+costheta))/sin theta=2cosectheta`
`tan theta /((1 - cot theta )) + cot theta /((1 - tan theta)) = (1+ sec theta cosec theta)`
(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to
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Prove that: `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(sin^2 A - cos^2 A)`.
Prove that sin6A + cos6A = 1 – 3sin2A . cos2A.
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.
