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प्रश्न
Prove that `(cos θ)/(1 - sin θ) = (1 + sin θ)/(cos θ)`.
उत्तर
L.H.S. = `cos θ/(1 - sin θ)`
= `(cos θ(1 + sin θ))/((1 - sin θ)(1 + sin θ))`
= `(cos θ(1 + sin θ))/(1 - sin^2θ)`
= `(cos θ(1 + sin θ))/(cos^2 θ)`
= `( 1 + sin θ)/cos θ`
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove that
`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
Prove the following identities:
`tan A - cot A = (1 - 2cos^2A)/(sin A cos A)`
Show that none of the following is an identity:
(i) `cos^2theta + cos theta =1`
If cosec2 θ (1 + cos θ) (1 − cos θ) = λ, then find the value of λ.
Prove the following identity :
`(1 - tanA)^2 + (1 + tanA)^2 = 2sec^2A`
Prove the following identity :
(secA - cosA)(secA + cosA) = `sin^2A + tan^2A`
Prove the following identity :
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Prove the following identity :
`2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1 = 0`
Prove the following identities.
`(sin^3"A" + cos^3"A")/(sin"A" + cos"A") + (sin^3"A" - cos^3"A")/(sin"A" - cos"A")` = 2
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= R.H.S
