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प्रश्न
Prove that sec2 (90° - θ) + tan2 (90° - θ) = 1 + 2 cot2 θ.
उत्तर
LHS = sec2 (90° - θ) + tan2 (90° - θ)
= cosec2θ + cot2θ
= 1 + cot2θ + cot2θ
= 1 + 2cot2θ
= RHS
Hence proved.
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संबंधित प्रश्न
Prove the following trigonometric identity.
`(sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta)`
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`cot^2A/(cosecA + 1)^2 = (1 - sinA)/(1 + sinA)`
Prove the following identities:
(1 + tan A + sec A) (1 + cot A – cosec A) = 2
If tanθ `= 3/4` then find the value of secθ.
Simplify : 2 sin30 + 3 tan45.
Prove the following identity :
secA(1 + sinA)(secA - tanA) = 1
Without using a trigonometric table, prove that
`(cos 70°)/(sin 20°) + (cos 59°)/(sin 31°) - 8sin^2 30° = 0`.
Prove the following identities.
`(sin "A" - sin "B")/(cos "A" + cos "B") + (cos "A" - cos "B")/(sin "A" + sin "B")`
Prove that `[(1 + sin theta - cos theta)/(1 + sin theta + cos theta)]^2 = (1 - cos theta)/(1 + cos theta)`
If tan θ = `9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ......[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
