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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 trigonometric identities.
`(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`
Prove the following trigonometric identity:
`sqrt((1 + sin A)/(1 - sin A)) = sec A + tan A`
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
`1 + (tan^2 θ)/((1 + sec θ)) = sec θ`
`tan theta/(1+ tan^2 theta)^2 + cottheta/(1+ cot^2 theta)^2 = sin theta cos theta`
Given `cos38^circ sec(90^circ - 2A) = 1` , Find the value of <A
If x = a sec θ + b tan θ and y = a tan θ + b sec θ prove that x2 - y2 = a2 - b2.
Prove that: 2(sin6θ + cos6θ) - 3 ( sin4θ + cos4θ) + 1 = 0.
Prove that : `1 - (cos^2 θ)/(1 + sin θ) = sin θ`.
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
