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प्रश्न
Prove that the following identities:
Sec A( 1 + sin A)( sec A - tan A) = 1.
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उत्तर
LHS = sec A(1 + sin A )( sec A - tan A)
= `1/cos A (1 + sin A) (1/cos A - sin A/cos A)`
= `1/cos A (1 + sin A) ((1 - sin A)/cos A)`
= `(1 - sin^2 A)/(cos^2 A) = (cos^2 A)/(cos^2 A)`
= 1
= RHS
Hence proved.
संबंधित प्रश्न
Prove that ` \frac{\sin \theta -\cos \theta +1}{\sin\theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta }` using the identity sec2 θ = 1 + tan2 θ.
Prove the following identities:
(cos A + sin A)2 + (cos A – sin A)2 = 2
Prove the following identities:
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2A * cos^2B)`
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Prove the following identities:
sec4 A (1 – sin4 A) – 2 tan2 A = 1
`cosec theta (1+costheta)(cosectheta - cot theta )=1`
If` (sec theta + tan theta)= m and ( sec theta - tan theta ) = n ,` show that mn =1
Prove the following identity :
`(1 - tanA)^2 + (1 + tanA)^2 = 2sec^2A`
Prove that :(sinθ+cosecθ)2+(cosθ+ secθ)2 = 7 + tan2 θ+cot2 θ.
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
