Advertisements
Advertisements
Question
Find the value of `θ(0^circ < θ < 90^circ)` if :
`cos 63^circ sec(90^circ - θ) = 1`
Advertisements
Solution
`cos63^circ sec(90^circ - θ) = 1`
`cos 63^circ cosecθ = 1`
⇒ `cos63^circ = sinθ`
⇒ `cos 63^circ = cos(90^circ - θ)`
⇒ `63^circ = 90^circ - θ`
⇒ `θ = 27^circ`
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities
sec4 A(1 − sin4 A) − 2 tan2 A = 1
Prove the following trigonometric identities.
`(cot^2 A(sec A - 1))/(1 + sin A) = sec^2 A ((1 - sin A)/(1 + sec A))`
if `cosec theta - sin theta = a^3`, `sec theta - cos theta = b^3` prove that `a^2 b^2 (a^2 + b^2) = 1`
Prove that: `sqrt((sec theta - 1)/(sec theta + 1)) + sqrt((sec theta + 1)/(sec theta - 1)) = 2 cosec theta`
Show that : `sinAcosA - (sinAcos(90^circ - A)cosA)/sec(90^circ - A) - (cosAsin(90^circ - A)sinA)/(cosec(90^circ - A)) = 0`
`(sec^2 theta-1) cot ^2 theta=1`
`(sec^2 theta -1)(cosec^2 theta - 1)=1`
Find the value of ( sin2 33° + sin2 57°).
Without using the trigonometric table, prove that
cos 1°cos 2°cos 3° ....cos 180° = 0.
`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`, then find the value of θ.
