Advertisements
Advertisements
प्रश्न
Prove the following identities:
sec2 A + cosec2 A = sec2 A . cosec2 A
उत्तर
L.H.S. = sec2 A + cosec2 A
= `1/(cos^2A) + 1/(sin^2A)`
= `(sin^2A + cos^2A)/(cos^2A sin^2A)`
= `1/(cos^2A sin^2A)`
= sec2 A cosec2 A
= R.H.S. ...(∵ sin2 A + cos2 A = 1)
APPEARS IN
संबंधित प्रश्न
If sinθ + sin2 θ = 1, prove that cos2 θ + cos4 θ = 1
Prove the following trigonometric identities.
sec6θ = tan6θ + 3 tan2θ sec2θ + 1
if `x/a cos theta + y/b sin theta = 1` and `x/a sin theta - y/b cos theta = 1` prove that `x^2/a^2 + y^2/b^2 = 2`
`cot theta/((cosec theta + 1) )+ ((cosec theta +1 ))/ cot theta = 2 sec theta `
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
If \[\sin \theta = \frac{1}{3}\] then find the value of 2cot2 θ + 2.
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
If `asin^2θ + bcos^2θ = c and p sin^2θ + qcos^2θ = r` , prove that (b - c)(r - p) = (c - a)(q - r)
Without using trigonometric table , evaluate :
`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
Without using trigonometric identity , show that :
`sin(50^circ + θ) - cos(40^circ - θ) = 0`