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Question
Prove the following identities:
`(1 + sin A)/(1 - sin A) = (cosec A + 1)/(cosec A - 1)`
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Solution
R.H.S = `(1/(sin A) + 1)/(1/(sin A) - 1)`
= `((1 + sin A)/sin A)/((1 - sin A)/sin A)`
= `((1 + sin A))/cancelsin A xx cancelsin A/((1 - sin A))`
= `(1 + sin A)/(1 - sin A)`
∴ R.H.S = L.H.S
RELATED QUESTIONS
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`(i) (sinθ + cosecθ)^2 + (cosθ + secθ)^2 = 7 + tan^2 θ + cot^2 θ`
`(ii) (sinθ + secθ)^2 + (cosθ + cosecθ)^2 = (1 + secθ cosecθ)^2`
`(iii) sec^4 θ– sec^2 θ = tan^4 θ + tan^2 θ`
(secA + tanA) (1 − sinA) = ______.
Prove the following trigonometric identity.
`(sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta)`
Prove the following trigonometric identities.
`1 + cot^2 theta/(1 + cosec theta) = cosec theta`
Prove that:
`tanA/(1 - cotA) + cotA/(1 - tanA) = secA "cosec" A + 1`
If `cosec theta = 2x and cot theta = 2/x ," find the value of" 2 ( x^2 - 1/ (x^2))`
Prove that:
`sqrt((sectheta - 1)/(sec theta + 1)) + sqrt((sectheta + 1)/(sectheta - 1)) = 2cosectheta`
Without using the trigonometric table, prove that
cos 1°cos 2°cos 3° ....cos 180° = 0.
tan θ × `sqrt(1 - sin^2 θ)` is equal to:
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`
