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प्रश्न
Prove the following identities:
`1/(cosA + sinA) + 1/(cosA - sinA) = (2cosA)/(2cos^2A - 1)`
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उत्तर १
`1/(cosA + sinA) + 1/(cosA - sinA)`
= `(cosA + sinA + cosA - sinA)/((cosA + sinA)(cosA - sinA)`
= `(2cosA)/(cos^2A - sin^2A)`
= `(2cosA)/(cos^2A - (1 - cos^2A))`
= `(2cosA)/(cos^2A - 1 + cos^2A)`
= `(2cosA)/(2cos^2A - 1)`
उत्तर २
`1/(cosA + sinA) + 1/(cosA - sinA)`
`1/(cosA + sinA) + 1/(cosA - sinA) = ((cos A - sin A) + (cos A + sin A))/((cos A + ain A)(cos A - sin A))`
(cosA − sinA) + (cosA + sinA) = 2 cosA
(cosA + sinA) (cosA − sinA) = cos2A − sin2A = cos(2A)
cos(2A) = 2cos2A − 1
`(2cosA)/cos(2A) = (2cosA)/(2cos^2 A-1)`
`1/(cosA + sinA) + 1/(cosA - sinA) = (2cosA)/(2cos^2A - 1)`
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + sec theta)/sec theta = (sin^2 theta)/(1 - cos theta)`
Show that : `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec A cosec A`
If `cos theta = 2/3 , " write the value of" (4+4 tan^2 theta).`
The value of sin ( \[{45}^° + \theta) - \cos ( {45}^°- \theta)\] is equal to
Prove the following identity :
`(1 - sin^2θ)sec^2θ = 1`
Prove the following identity :
`(secθ - tanθ)^2 = (1 - sinθ)/(1 + sinθ)`
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
Prove that cosec θ – cot θ = `sin theta/(1 + cos theta)`
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`
