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प्रश्न
If A + B = 90°, show that sec2 A + sec2 B = sec2 A. sec2 B.
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उत्तर
LHS = sec2A + sec2B
= `1/(cos^2A) + 1/(cos^2 B)`
= `1/(cos^2A) + 1/(cos^2(90° - A))`
= `1/(cos^2 A) + 1/(sin^2 A)`
= `1/( sin^ A. cos^2 A)`
= sec2 A cosec2 A
= sec2 A cosec2 (90° - B)
= sec2A. sec2 B = RHS
Hence proved.
संबंधित प्रश्न
Show that `sqrt((1+cosA)/(1-cosA)) = cosec A + cot A`
Prove that `cosA/(1+sinA) + tan A = secA`
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = cosA/(1 - sinA)`
Prove the following identity :
`cosA/(1 - tanA) + sin^2A/(sinA - cosA) = cosA + sinA`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`tan35^circ cot(90^circ - θ) = 1`
Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.
Prove that `(sin 70°)/(cos 20°) + (cosec 20°)/(sec 70°) - 2 cos 70° xx cosec 20°` = 0.
Prove that `sqrt(sec^2 theta + "cosec"^2 theta) = tan theta + cot theta`
Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
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`
