Advertisements
Advertisements
प्रश्न
Prove that `(tan^2 theta - 1)/(tan^2 theta + 1)` = 1 – 2 cos2θ
Advertisements
उत्तर
L.H.S = `(tan^2 theta - 1)/(tan^2 theta + 1)`
= `(tan^2 theta - 1)/(sec^2 theta)`
= `(sin^2 theta)/(cos^2 theta) - 1 ÷ 1/(cos^2 theta)`
= `(sin^2 theta - cos^2 theta)/(cos^2 theta) xx (cos^2 theta)/1`
= sin2θ − cos2θ
= 1 − cos2θ − cos2θ
= 1 – cos2θ
L.H.S = R.H.S
Hence it is proved.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
tan2 A sec2 B − sec2 A tan2 B = tan2 A − tan2 B
If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ – 3 cos θ = ± 3.
Prove the following identities:
`(sinAtanA)/(1 - cosA) = 1 + secA`
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
Write the value of cos1° cos 2°........cos180° .
From the figure find the value of sinθ.
If cosec θ = 2x and \[5\left( x^2 - \frac{1}{x^2} \right)\] \[2\left( x^2 - \frac{1}{x^2} \right)\]
Prove the following identity :
`cosA/(1 - tanA) + sin^2A/(sinA - cosA) = cosA + sinA`
Express (sin 67° + cos 75°) in terms of trigonometric ratios of the angle between 0° and 45°.
If cot θ + tan θ = x and sec θ – cos θ = y, then prove that `(x^2y)^(2/3) – (xy^2)^(2/3)` = 1
