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प्रश्न
Prove the following identities: sec2 θ + cosec2 θ = sec2 θ cosec2 θ.
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उत्तर
LHS = sec2 θ + cosec2 θ
= `1/(cos^2 θ) + 1/(sin^2 θ)`
= `(sin^2 θ + cos^2 θ)/(sin^2 θ. cos^2 θ)`
= `1/(sin^2 θ. cos^2 θ)`
= `1/(sin^2 θ) xx 1/(cos^2 θ)`
= sec2 θ cosec2 θ
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
sec A (1 − sin A) (sec A + tan A) = 1
Prove the following trigonometric identities.
`sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A + 1) = 1`
Prove the following identities:
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
What is the value of (1 + cot2 θ) sin2 θ?
Find the value of sin 30° + cos 60°.
Express (sin 67° + cos 75°) in terms of trigonometric ratios of the angle between 0° and 45°.
If x sin3θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ , then show that x2 + y2 = 1.
Choose the correct alternative:
sec 60° = ?
`5/(sin^2theta) - 5cot^2theta`, complete the activity given below.
Activity:
`5/(sin^2theta) - 5cot^2theta`
= `square (1/(sin^2theta) - cot^2theta)`
= `5(square - cot^2theta) ......[1/(sin^2theta) = square]`
= 5(1)
= `square`
Prove the following:
`sintheta/(1 + cos theta) + (1 + cos theta)/sintheta` = 2cosecθ
