Advertisements
Advertisements
प्रश्न
If 1 – cos2θ = `1/4`, then θ = ?
उत्तर
1 – cos2θ = `1/4` ......[Given]
∴ sin2θ = `1/4` .....`[(because sin^2theta + cos^2theta = 1),(therefore 1 - cos^2theta = sin^2theta)]`
∴ sin θ = `1/2` ......[Taking square root of both sides]
∴ θ = 30° ......`[because sin 30^circ = 1/2]`
APPEARS IN
संबंधित प्रश्न
Evaluate
`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)`
(1 + tan θ + sec θ) (1 + cot θ − cosec θ) = ______.
Evaluate without using trigonometric tables:
`cos^2 26^@ + cos 64^@ sin 26^@ + (tan 36^@)/(cot 54^@)`
`(sec theta -1 )/( sec theta +1) = ( sin ^2 theta)/( (1+ cos theta )^2)`
If x=a `cos^3 theta and y = b sin ^3 theta ," prove that " (x/a)^(2/3) + ( y/b)^(2/3) = 1.`
Write the value of `(1 + tan^2 theta ) cos^2 theta`.
If `cos theta = 2/3 , " write the value of" (4+4 tan^2 theta).`
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
Prove the following identity :
`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`
Prove the following identity :
`(tanθ + sinθ)/(tanθ - sinθ) = (secθ + 1)/(secθ - 1)`
If `x/(a cosθ) = y/(b sinθ) "and" (ax)/cosθ - (by)/sinθ = a^2 - b^2 , "prove that" x^2/a^2 + y^2/b^2 = 1`
Prove that `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ = 1/(sin^2 θ. cos^2 θ) - 2`.
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= R.H.S
If sec θ = `41/40`, then find values of sin θ, cot θ, cosec θ
Prove that `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ
Prove that `sec"A"/(tan "A" + cot "A")` = sin A
Prove that sin6A + cos6A = 1 – 3sin2A . cos2A
Show that tan4θ + tan2θ = sec4θ – sec2θ.
If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 – 1) = 2p.
sec θ when expressed in term of cot θ, is equal to ______.
