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प्रश्न
Prove the following identity :
`(secθ - tanθ)^2 = (1 - sinθ)/(1 + sinθ)`
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उत्तर
`(secθ - tanθ)^2`
= `(1/cosθ - sinθ/cosθ)^2`
= `((1 - sinθ)/cosθ)^2 = (1 - sinθ)^2/cos^2θ`
= `(1 - sinθ)^2/(1 - sin^2θ) = (1 - sinθ)^2/((1 -sinθ)(1 + sinθ))` (∵ `1 - sin^2θ = cos^2θ`
= `(1 - sinθ)/(1 + sinθ)`
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संबंधित प्रश्न
If `sec alpha=2/sqrt3` , then find the value of `(1-cosecalpha)/(1+cosecalpha)` where α is in IV quadrant.
Prove that:
(sec A − tan A)2 (1 + sin A) = (1 − sin A)
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
cosec4 θ − cosec2 θ = cot4 θ + cot2 θ
Prove the following identity :
`(1 - sin^2θ)sec^2θ = 1`
Prove the following identity :
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
Prove the following identity :
`sec^4A - sec^2A = sin^2A/cos^4A`
Prove that : `1 - (cos^2 θ)/(1 + sin θ) = sin θ`.
If `cos theta/(1 + sin theta) = 1/"a"`, then prove that `("a"^2 - 1)/("a"^2 + 1)` = sin θ
Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
