Advertisements
Advertisements
प्रश्न
Show that : `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec A cosec A`
Advertisements
उत्तर
L.H.S. = `sinA/(sin (90^circ - A)) + cosA/(cos(90^circ - A))`
= `sinA/cosA + cosA/sinA`
= `(sin^2A + cos^2A)/(cosAsinA)` ...(∵ sin2 A + cos2 A = 1)
= `1/(cosAsinA)`
= sec A cosec A = R.H.S.
APPEARS IN
संबंधित प्रश्न
Evaluate sin25° cos65° + cos25° sin65°
`(1+tan^2A)/(1+cot^2A)` = ______.
Prove the following identities:
`(1 + sin A)/(1 - sin A) = (cosec A + 1)/(cosec A - 1)`
If sin A + cos A = p and sec A + cosec A = q, then prove that : q(p2 – 1) = 2p.
If 5x = sec ` theta and 5/x = tan theta , " find the value of 5 "( x^2 - 1/( x^2))`
If cosec θ − cot θ = α, write the value of cosec θ + cot α.
Prove that `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec(90^circ - A) cosec(90^circ - A)`
Prove that `(cos θ)/(1 - sin θ) = (1 + sin θ)/(cos θ)`.
Prove that `( tan A + sec A - 1)/(tan A - sec A + 1) = (1 + sin A)/cos A`.
Prove the following that:
`tan^3θ/(1 + tan^2θ) + cot^3θ/(1 + cot^2θ)` = secθ cosecθ – 2 sinθ cosθ
