Advertisements
Advertisements
प्रश्न
Find x , if `cos(2x - 6) = cos^2 30^circ - cos^2 60^circ`
Advertisements
उत्तर
`cos(2x - 6) = cos^2 30^circ - cos^2 60^circ`
⇒ `cos(2x - 6) = cos^2 (90^circ - 60^circ) - cos^2 60^circ`
⇒ `cos(2x - 6) = sin^2 60^circ - cos^2 60^circ`
⇒ `cos(2x - 6) = 1 - 2cos^2 60^circ = 1 - 2(1/2)^2 = 1 - 1/2 = 1/2`
⇒ `cos(2x - 6) = 1/2`
⇒ `cos(2x - 6) = cos60^circ`
⇒ `(2x - 6) = 60^circ`
⇒ `2x = 66^circ`
⇒ `x = 33^circ`
APPEARS IN
संबंधित प्रश्न
Express the ratios cos A, tan A and sec A in terms of sin A.
Prove the following trigonometric identities
tan2 A + cot2 A = sec2 A cosec2 A − 2
Prove the following trigonometric identities.
if `T_n = sin^n theta + cos^n theta`, prove that `(T_3 - T_5)/T_1 = (T_5 - T_7)/T_3`
`(1+tan^2theta)(1+cot^2 theta)=1/((sin^2 theta- sin^4theta))`
Write the value of `(sin^2 theta 1/(1+tan^2 theta))`.
`If sin theta = cos( theta - 45° ),where theta " is acute, find the value of "theta` .
If cos A + cos2 A = 1, then sin2 A + sin4 A =
Prove that cosec2 (90° - θ) + cot2 (90° - θ) = 1 + 2 tan2 θ.
Prove the following identities.
`costheta/(1 + sintheta)` = sec θ – tan θ
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
