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.
संबंधित प्रश्न
Prove the following trigonometric identities.
`cot theta - tan theta = (2 cos^2 theta - 1)/(sin theta cos theta)`
Prove the following trigonometric identities.
`(cos^2 theta)/sin theta - cosec theta + sin theta = 0`
If `sec theta = x ,"write the value of tan" theta`.
If sec θ + tan θ = x, then sec θ =
Prove the following identity :
( 1 + cotθ - cosecθ) ( 1 + tanθ + secθ)
Prove the following identities:
`(sec"A"-1)/(sec"A"+1)=(sin"A"/(1+cos"A"))^2`
Prove that `(cot "A" + "cosec A" - 1)/(cot "A" - "cosec A" + 1) = (1 + cos "A")/sin "A"`
Prove that cot2θ – tan2θ = cosec2θ – sec2θ
Prove that `sqrt(sec^2 theta + "cosec"^2 theta) = tan theta + cot theta`
Prove the following:
(sin α + cos α)(tan α + cot α) = sec α + cosec α
