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प्रश्न
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
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उत्तर
LHS = sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ )
= sec θ. sec θ - tan θ. tan θ
= sec2θ - tan2θ
= 1
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + cos A)/sin A = sin A/(1 - cos A)`
Prove the following trigonometric identities.
(sec A − cosec A) (1 + tan A + cot A) = tan A sec A − cot A cosec A
Prove the following identities:
`cot^2A/(cosecA + 1)^2 = (1 - sinA)/(1 + sinA)`
Prove the following identities:
`(cos theta "cosec" theta - sin theta sec theta )/(cos theta + sin theta) = "cosec" theta - sec theta`
Write the value of ` cosec^2 (90°- theta ) - tan^2 theta`
(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, then\[\frac{x^2}{a^2} + \frac{y^2}{b^2}\]
Prove that `tan A/(1 + tan^2 A)^2 + cot A/(1 + cot^2 A)^2 = sin A.cos A`
Prove that : `tan"A"/(1 - cot"A") + cot"A"/(1 - tan"A") = sec"A".cosec"A" + 1`.
Prove that `(sin^2θ)/(cos θ) + cos θ = sec θ`.
