Advertisements
Advertisements
Question
Show that : `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec A cosec A`
Advertisements
Solution
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
RELATED QUESTIONS
Prove the following trigonometric identities.
`(cosec A)/(cosec A - 1) + (cosec A)/(cosec A = 1) = 2 sec^2 A`
Prove the following trigonometric identity.
`(sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta)`
Prove the following identities:
`1/(tan A + cot A) = cos A sin A`
Prove the following identity :
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
Prove that: (1+cot A - cosecA)(1 + tan A+ secA) =2.
Prove the following identities:
`(1 - tan^2 θ)/(cot^2 θ - 1) = tan^2 θ`.
If `sqrt(3)` sin θ – cos θ = θ, then show that tan 3θ = `(3tan theta - tan^3 theta)/(1 - 3 tan^2 theta)`
Which is not correct formula?
`(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = ?
Prove that sec2θ + cosec2θ = sec2θ × cosec2θ.
