Advertisements
Advertisements
प्रश्न
Prove that `((1 + sin θ - cos θ)/( 1 + sin θ + cos θ))^2 = (1 - cos θ)/(1 + cos θ)`.
Advertisements
उत्तर
LHS = `((1 + sin θ - cos θ)/( 1 + sin θ + cos θ))^2`
⇒ `(1 + sin^2 θ + cos^2 θ + 2(sin θ - cos θ - sin θ. cos θ))/(1 + sin^2 θ + cos^2 θ + 2(sin θ + cos θ + sin θ. cos θ)`
= `(1 + 1 + 2 (sin θ - cos θ - sin θ. cos θ))/( 1 + 1 + 2((sin θ + cos θ + sin θ. cos θ)`
= `(2 (1 + sin θ - cos θ - sin θ. cos θ))/(2( 1 + (sin θ + cos θ + sin θ. cos θ))`
= `( 1 + sin θ - cos θ( 1 + sin θ))/(1 + sin θ + cos θ( 1 + sin θ))`
= `((1 + sin θ)(1 - cos θ))/((1 + sin θ)( 1 + cos θ))`
= `(1 - cos θ)/( 1 + cos θ)`
= RHS
Hence proved.
संबंधित प्रश्न
Evaluate
`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)`
Prove the following trigonometric identities.
sec A (1 − sin A) (sec A + tan A) = 1
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that a2 + b2 = m2 + n2
If sin A + cos A = m and sec A + cosec A = n, show that : n (m2 – 1) = 2 m
`(1+ tan^2 theta)/(1+ tan^2 theta)= (cos^2 theta - sin^2 theta)`
`(sec theta + tan theta )/( sec theta - tan theta ) = ( sec theta + tan theta )^2 = 1+2 tan^2 theta + 25 sec theta tan theta `
If x sin3θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ , then show that x2 + y2 = 1.
If A + B = 90°, show that sec2 A + sec2 B = sec2 A. sec2 B.
Prove the following identities.
(sin θ + sec θ)2 + (cos θ + cosec θ)2 = 1 + (sec θ + cosec θ)2
Prove that `(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2
