Advertisements
Advertisements
प्रश्न
If A + B = 90°, show that `(sin B + cos A)/sin A = 2tan B + tan A.`
Advertisements
उत्तर
LHS = `(sin B + sec A)/sin A`
= `(sin (90 - A) + sec A)/sin A`
= `(cos A + sec A)/sin A`
= `(cos^2 A + 1)/(sin A. cos A)`
= `(2cos^2 A + sin^2 A)/(sin A. cos A)`
= 2cot A + tan A
= 2 tan B + tan A = RHS
hence proved.
संबंधित प्रश्न
If cos θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1
Prove the following identities:
`(sinAtanA)/(1 - cosA) = 1 + secA`
Prove the following identities:
`((cosecA - cotA)^2 + 1)/(secA(cosecA - cotA)) = 2cotA`
If 2 sin A – 1 = 0, show that: sin 3A = 3 sin A – 4 sin3 A
What is the value of (1 − cos2 θ) cosec2 θ?
Prove that `sin(90^circ - A).cos(90^circ - A) = tanA/(1 + tan^2A)`
`(1 + cot^2A)/(1 + tan^2A)` = ?
Prove that sec2θ + cosec2θ = sec2θ × cosec2θ.
Prove that `(cos^2θ)/(sinθ) + sin θ = "cosec" θ`.
If cosec A – sin A = p and sec A – cos A = q, then prove that `(p^2q)^(2/3) + (pq^2)^(2/3) = 1`.
