Advertisements
Advertisements
Question
`1/(1 - sinθ) + 1/(1 + sinθ)` = 2sec2θ
Advertisements
Solution
डावी बाजू = `1/(1 - sinθ) + 1/(1 + sinθ)`
= `((1 + sinθ) + (1 - sinθ))/((1 - sinθ)(1 + sinθ))`
= `(1 + sinθ + 1 - sinθ)/((1 - sinθ)(1 + sinθ))`
= `2/(1 - sin^2θ)`
= `2/cos^2θ` .....[∵ 1 - sin2θ = cos2θ]
= `2 xx 1/cos^2θ`
= 2sec2θ
= उजवी बाजू
∴ `1/(1 - sinθ) + 1/(1 + sinθ)` = 2sec2θ
APPEARS IN
RELATED QUESTIONS
sec θ(1 - sin θ) (sec θ + tan θ) = 1
`(sin^2theta)/(cos theta) + cos theta` = sec θ हे सिद्ध करा.
जर tan θ + cot θ = 2, तर tan2θ + cot2θ = ?
cot2θ × sec2θ = cot2θ + 1 हे सिद्ध करा.
जर 3 sin θ = 4 cos θ, तर sec θ = ?
`(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2 हे सिद्ध करा.
`(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B हे सिद्ध करा.
sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")` हे सिद्ध करा.
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"` हे सिद्ध करा.
जर cosec A – sin A = p आणि sec A – cos A = q, तर सिद्ध करा. `("p"^2"q")^(2/3) + ("pq"^2)^(2/3)` = 1
