Advertisements
Advertisements
प्रश्न
Find the value of x , if `cosx = cos60^circ cos30^circ - sin60^circ sin30^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
If cos θ + cos2 θ = 1, prove that sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1
Prove the following identities:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Prove the following identities:
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (1 + cosA)/sinA`
Prove the following identities:
(1 + tan A + sec A) (1 + cot A – cosec A) = 2
Prove that:
`"tan A"/(1 + "tan"^2 "A")^2 + "Cot A"/(1 + "Cot"^2 "A")^2 = "sin A cos A"`.
Prove the following identity :
secA(1 + sinA)(secA - tanA) = 1
Prove that cot2θ – tan2θ = cosec2θ – sec2θ
Prove that `(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B
Simplify (1 + tan2θ)(1 – sinθ)(1 + sinθ)
Show that: `tan "A"/(1 + tan^2 "A")^2 + cot "A"/(1 + cot^2 "A")^2 = sin"A" xx cos"A"`
