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प्रश्न
Prove that : `(sin(90° - θ) tan(90° - θ) sec (90° - θ))/(cosec θ. cos θ. cot θ) = 1`
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उत्तर
LHS = `(sin(90° - θ) tan(90° - θ) sec (90° - θ))/(cosec θ. cos θ. cot θ) = 1`
= `(cosec θ. cos θ. cot θ)/(cosec θ. cos θ. cot θ)`
= 1
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`sin^2 A + 1/(1 + tan^2 A) = 1`
Prove that:
(sin A + cos A) (sec A + cosec A) = 2 + sec A cosec A
`1+(tan^2 theta)/((1+ sec theta))= sec theta`
If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.
Prove the following identity :
`(1 - tanA)^2 + (1 + tanA)^2 = 2sec^2A`
If cosθ + sinθ = `sqrt2` cosθ, show that cosθ - sinθ = `sqrt2` sinθ.
If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, then prove that x2 + y2 = 1
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
The value of 2sinθ can be `a + 1/a`, where a is a positive number, and a ≠ 1.
If `sqrt(3) tan θ` = 1, then find the value of sin2θ – cos2θ.
