Advertisements
Advertisements
प्रश्न
Prove the following identities:
`1/(sin θ + cos θ) + 1/(sin θ - cos θ) = (2sin θ)/(1 - 2 cos^2 θ)`.
Advertisements
उत्तर
LHS = `1/(sin θ + cos θ) + 1/(sin θ - cos θ)`
= `((sin θ - cos θ) + (sin θ + cos θ))/(sin^2 θ - cos^2 θ)`
= `(2 sin θ)/((1 - cos^2 θ) - cos^2 θ)`
= `(2 sin θ)/(1 - 2cos^2 θ)`
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + cot A + tan A)(sin A - cos A) = sec A/(cosec^2 A) - (cosec A)/sec^2 A = sin A tan A - cos A cot A`
Prove the following trigonometric identities.
sin2 A cos2 B − cos2 A sin2 B = sin2 A − sin2 B
If 4 cos2 A – 3 = 0, show that: cos 3 A = 4 cos3 A – 3 cos A
cosec4 θ − cosec2 θ = cot4 θ + cot2 θ
Write True' or False' and justify your answer the following :
The value of \[\sin \theta\] is \[x + \frac{1}{x}\] where 'x' is a positive real number .
Prove the following identity :
`(secA - 1)/(secA + 1) = sin^2A/(1 + cosA)^2`
Express (sin 67° + cos 75°) in terms of trigonometric ratios of the angle between 0° and 45°.
If A + B = 90°, show that `(sin B + cos A)/sin A = 2tan B + tan A.`
Prove that sin6A + cos6A = 1 – 3sin2A . cos2A.
The value of 2sinθ can be `a + 1/a`, where a is a positive number, and a ≠ 1.
