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प्रश्न
Prove that:
`1/(sinA - cosA) - 1/(sinA + cosA) = (2cosA)/(2sin^2A - 1)`
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उत्तर
`1/(sinA - cosA) - 1/(sinA + cosA)`
= `(sinA + cosA - sinA + cosA)/((sinA - cosA)(sinA + cosA)`
= `(2cosA)/(sin^2A - cos^2A)`
= `(2cosA)/(sin^2A - (1 - sin^2A))`
= `(2cosA)/(sin^2A - 1 + sin^2A)`
= `(2cosA)/(2sin^2A - 1)`
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संबंधित प्रश्न
Prove that (1 + cot θ – cosec θ)(1+ tan θ + sec θ) = 2
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(1 + cot2 A) sin2 A = 1
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`(cos^2 theta)/sin theta - cosec theta + sin theta = 0`
Prove the following trigonometric identities.
`cos A/(1 - tan A) + sin A/(1 - cot A) = sin A + cos A`
Prove that `( sintheta - 2 sin ^3 theta ) = ( 2 cos ^3 theta - cos theta) tan theta`
Prove the following identity :
`(1 + cotA)^2 + (1 - cotA)^2 = 2cosec^2A`
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
If sin θ + sin2 θ = 1 show that: cos2 θ + cos4 θ = 1
The value of 2sinθ can be `a + 1/a`, where a is a positive number, and a ≠ 1.
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.
