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प्रश्न
Prove the following trigonometric identities.
`(1 + cos θ + sin θ)/(1 + cos θ - sin θ) = (1 + sin θ)/cos θ`
Prove the following:
`(1 + cos θ + sin θ)/(1 + cos θ - sin θ) = (1 + sin θ)/cos θ`
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उत्तर
`(1 + cos θ + sin θ)/(1 + cos θ - sin θ) = (1 + sin θ)/cos θ`
Consider the LHS = `(1 + cos θ + sin θ)/(1 + cos θ - sin θ)`
`= ((1 + cos θ + sin θ)/(1 + cos θ - sin θ))((1 + cos θ + sin θ)/(1 + cos θ + sin θ))`
`= (1 + cos θ + sin θ)^2/((1 + cos θ)^2 sin^2 θ)`
`= (2 + 2(cos θ + sin θ + sin θ cos θ))/(2 cos^2 θ+ 2 cos θ)`
`= (2(1 + cos θ)(1 + sin θ))/(2 cos θ (1 + cos θ))`
`= (1 + sin θ)/cos θ`
= RHS
Hence proved
संबंधित प्रश्न
Prove the following identities:
cosec A(1 + cos A) (cosec A – cot A) = 1
If cosec2 θ (1 + cos θ) (1 − cos θ) = λ, then find the value of λ.
sec4 A − sec2 A is equal to
Prove the following identity :
`sin^8θ - cos^8θ = (sin^2θ - cos^2θ)(1 - 2sin^2θcos^2θ)`
If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, then prove that x2 + y2 = 1
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 = 41/40`, then find values of sin θ, cot θ, cosec θ.
Prove that `(1 + sin B)/(cos B) + (cos B)/(1 + sin B) = 2 sec B`.
Prove that (sec θ + tan θ) (1 – sin θ) = cos 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`
