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प्रश्न
tan θ × `sqrt(1 - sin^2 θ)` is equal to:
पर्याय
cos θ
sin θ
tan θ
cot θ
उत्तर
sin θ
Explanation:
`tan θ xx sqrt(1 - sin^2 θ) ...{sin^2 θ + cos^2 θ = 1, ∴ cos^2 θ = 1 - sin^2 θ}`
= `tan θ xx sqrt(cos^2 θ)`
= tan θ × cos θ
= `(sin θ)/(cos θ)` × cos θ
= sin θ
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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)`
Prove the following identities:
(cosec A + sin A) (cosec A – sin A) = cot2 A + cos2 A
Prove that:
(tan A + cot A) (cosec A – sin A) (sec A – cos A) = 1
`(1+tan^2theta)(1+cot^2 theta)=1/((sin^2 theta- sin^4theta))`
What is the value of \[\sin^2 \theta + \frac{1}{1 + \tan^2 \theta}\]
If cosec θ = 2x and \[5\left( x^2 - \frac{1}{x^2} \right)\] \[2\left( x^2 - \frac{1}{x^2} \right)\]
Prove the following identity :
`(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ) = 2(1 + cotθ)`
If a cos θ – b sin θ = c, then prove that (a sin θ + b cos θ) = `± sqrt("a"^2 + "b"^2 -"c"^2)`
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)
If tan θ + cot θ = 2, then tan2θ + cot2θ = ?
