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प्रश्न
Prove that the following identities:
Sec A( 1 + sin A)( sec A - tan A) = 1.
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.
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संबंधित प्रश्न
Prove the following trigonometric identities.
`sqrt((1 - cos A)/(1 + cos A)) = cosec A - cot A`
Prove the following identities:
cot2 A – cos2 A = cos2 A . cot2 A
Prove that:
(1 + tan A . tan B)2 + (tan A – tan B)2 = sec2 A sec2 B
`(sin theta+1-cos theta)/(cos theta-1+sin theta) = (1+ sin theta)/(cos theta)`
`If sin theta = cos( theta - 45° ),where theta " is acute, find the value of "theta` .
The value of \[\sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}}\]
Without using trigonometric identity , show that :
`sin(50^circ + θ) - cos(40^circ - θ) = 0`
Prove that : `1 - (cos^2 θ)/(1 + sin θ) = sin θ`.
Prove that `sqrt((1 + sin θ)/(1 - sin θ))` = sec θ + tan θ.
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
