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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.
`(1/(sec^2 theta - cos theta) + 1/(cosec^2 theta - sin^2 theta)) sin^2 theta cos^2 theta = (1 - sin^2 theta cos^2 theta)/(2 + sin^2 theta + cos^2 theta)`
Write the value of sin A cos (90° − A) + cos A sin (90° − A).
Simplify
sin A `[[sinA -cosA],["cos A" " sinA"]] + cos A[[ cos A" sin A " ],[-sin A" cos A"]]`
Prove the following identity :
`sinA/(1 + cosA) + (1 + cosA)/sinA = 2cosecA`
Prove the following Identities :
`(cosecA)/(cotA+tanA)=cosA`
Prove the following identity :
`sqrt((1 + cosA)/(1 - cosA)) = cosecA + cotA`
Prove the following identity :
`(1 + tan^2θ)sinθcosθ = tanθ`
If A = 30°, verify that `sin 2A = (2 tan A)/(1 + tan^2 A)`.
tan θ × `sqrt(1 - sin^2 θ)` is equal to:
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`
