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प्रश्न
Prove that cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1 = 0.
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उत्तर
LHS = cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1
= cot θ. cot θ - cosec θ. cosec θ + 1
= (cot2θ - cosec2θ) + 1
= - 1 + 1 = 0
= RHS
Hence proved.
संबंधित प्रश्न
Prove the identity (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.
Prove the following trigonometric identities.
(sec2 θ − 1) (cosec2 θ − 1) = 1
Show that : `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec A cosec A`
Write the value of `cosec^2 theta (1+ cos theta ) (1- cos theta).`
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)
Choose the correct alternative:
cos 45° = ?
Prove that `costheta/(1 + sintheta) = (1 - sintheta)/(costheta)`
If `sqrt(3) tan θ` = 1, then find the value of sin2θ – cos2θ.
Show that tan4θ + tan2θ = sec4θ – sec2θ.
`(cos^2 θ)/(sin^2 θ) - 1/(sin^2 θ)`, in simplified form, is ______.
