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प्रश्न
Prove that `[(1 + sin theta - cos theta)/(1 + sin theta + cos theta)]^2 = (1 - cos theta)/(1 + cos theta)`
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उत्तर
L.H.S = `[(1 + sin theta - cos theta)/(1 + sin theta + cos theta)]^2`
= `(1 + sin^2theta + cos^2theta + 2sintheta - 2sintheta cos theta - 2costheta)/(1 + sin^2theta + cos^2theta + 2sintheta + 2sintheta costheta + 2costheta)`
= `(1 + 1 + 2sintheta (1 - cos theta) - 2cos theta)/(1 + 1 + 2sin theta + 2cos theta (sin theta + 1))`
= `(2(1 - cos theta) + 2sintheta (1 - cos theta))/(2(1 + sin theta) + 2cos theta(1 + sin theta))`
= `(2(1 - costheta)(1 + sintheta))/(2(1 + sintheta)(1 + costheta))`
= `((1 - cos theta))/((1 + cos theta))`
L.H.S = R.H.S
Hence it is proved.
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संबंधित प्रश्न
9 sec2 A − 9 tan2 A = ______.
Prove the following trigonometric identities.
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`(cosec A)/(cosec A - 1) + (cosec A)/(cosec A = 1) = 2 sec^2 A`
Prove the following identities:
(1 + tan A + sec A) (1 + cot A – cosec A) = 2
`cos^2 theta /((1 tan theta))+ sin ^3 theta/((sin theta - cos theta))=(1+sin theta cos theta)`
If `sqrt(3) sin theta = cos theta and theta ` is an acute angle, find the value of θ .
cot θ . tan θ = ?
(sec θ + tan θ) . (sec θ – tan θ) = ?
If tan θ + cot θ = 2, then tan2θ + cot2θ = ?
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
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`square/square` = cosec2θ ......[Taking root on the both side]
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and sin θ = `1/("cosec" θ)`
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∴ sin θ = `9/41`
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