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प्रश्न
Find x , if `cos(2x - 6) = cos^2 30^circ - cos^2 60^circ`
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उत्तर
`cos(2x - 6) = cos^2 30^circ - cos^2 60^circ`
⇒ `cos(2x - 6) = cos^2 (90^circ - 60^circ) - cos^2 60^circ`
⇒ `cos(2x - 6) = sin^2 60^circ - cos^2 60^circ`
⇒ `cos(2x - 6) = 1 - 2cos^2 60^circ = 1 - 2(1/2)^2 = 1 - 1/2 = 1/2`
⇒ `cos(2x - 6) = 1/2`
⇒ `cos(2x - 6) = cos60^circ`
⇒ `(2x - 6) = 60^circ`
⇒ `2x = 66^circ`
⇒ `x = 33^circ`
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संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(cosec θ – cot θ)^2 = (1-cos theta)/(1 + cos theta)`
`sin^2 theta + 1/((1+tan^2 theta))=1`
`(cos ec^theta + cot theta )/( cos ec theta - cot theta ) = (cosec theta + cot theta )^2 = 1+2 cot^2 theta + 2cosec theta cot theta`
\[\frac{x^2 - 1}{2x}\] is equal to
Prove that `(sec θ - 1)/(sec θ + 1) = ((sin θ)/(1 + cos θ ))^2`
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
Prove the following identities.
`costheta/(1 + sintheta)` = sec θ – tan θ
If `sqrt(3)` sin θ – cos θ = θ, then show that tan 3θ = `(3tan theta - tan^3 theta)/(1 - 3 tan^2 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)
