Advertisements
Advertisements
प्रश्न
Prove that `sin(90^circ - A).cos(90^circ - A) = tanA/(1 + tan^2A)`
उत्तर
LHS = `sin(90^circ - A).cos(90^circ - A)`
⇒ cosA.sinA
RHS = `tanA/(1 + tan^2A) = tanA/sec^2A = (sinA/cosA)/(1/cos^2A)`
⇒ RHS = `sinA/cosA . cos^2A = cosA.sinA`
Thus , LHS = RHS
⇒ `sin(90^circ - A).cos(90^circ - A) = tanA/(1 + tan^2A)`
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`sin^2 A + 1/(1 + tan^2 A) = 1`
Find the value of ` ( sin 50°)/(cos 40°)+ (cosec 40°)/(sec 50°) - 4 cos 50° cosec 40 °`
If `secθ = 25/7 ` then find tanθ.
Prove the following identity :
`sqrt((secq - 1)/(secq + 1)) + sqrt((secq + 1)/(secq - 1))` = 2 cosesq
Prove the following identity :
`(1 + cotA)^2 + (1 - cotA)^2 = 2cosec^2A`
Without using trigonometric table , evaluate :
`cosec49°cos41° + (tan31°)/(cot59°)`
Without using trigonometric identity , show that :
`sec70^circ sin20^circ - cos20^circ cosec70^circ = 0`
Prove that `(sin^2theta)/(cos theta) + cos theta` = sec θ
If tan θ + sec θ = l, then prove that sec θ = `(l^2 + 1)/(2l)`.
(1 + sin A)(1 – sin A) is equal to ______.
