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प्रश्न
Prove that `( 1 + sin θ)/(1 - sin θ) = 1 + 2 tan θ/cos θ + 2 tan^2 θ` .
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उत्तर
RHS = `1 + 2 tan θ/cos θ + 2 tan^2 θ`
= `1 + 2 sin θ/cos^2θ + 2 sin^2 θ/cos^2 θ`
= `(cos^2 θ + 2sin θ + 2 sin^2 θ)/(cos^2θ)`
= `(1 - sin^2θ + 2 sin θ + 2 sin^2θ )/(1 - sin^2θ)`
= `(1 + sin^2θ + 2 sin θ)/(1 - sin^2θ)`
= `(1 + sin θ)^2/( 1 + sin θ)(1 - sin θ)`
= `(1 + sin θ)/(1 - sin θ)`
= LHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
tan2θ cos2θ = 1 − cos2θ
Prove the following trigonometric identities.
`(cot^2 A(sec A - 1))/(1 + sin A) = sec^2 A ((1 - sin A)/(1 + sec A))`
`(sin theta)/((sec theta + tan theta -1)) + cos theta/((cosec theta + cot theta -1))=1`
If `m = (cos θ - sin θ)` and `n = (cos θ + sin θ)`, show that `sqrt(m/n) + sqrt(n/m) = 2/sqrt(1 - tan^2θ)`.
If sin θ − cos θ = 0 then the value of sin4θ + cos4θ
Prove the following identity :
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
Prove that: `sqrt((1 - cos θ)/(1 + cos θ)) = "cosec" θ - cot θ`.
Prove that:
`(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(2 sin^2 A - 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)
If `tan θ = 7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ...[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` ...`[cos theta = 1/sectheta]`
