Advertisements
Advertisements
प्रश्न
If 4 cos2 A – 3 = 0, show that: cos 3 A = 4 cos3 A – 3 cos A
उत्तर
4 cos2 A – 3 = 0
`=>` 4 cos2 A = 3
`=> cos^2A = 3/4`
`=> cosA = sqrt3/2`
We know cos 30° `= sqrt(3)/2`
So, A = 30°
L.H.S. = cos3 A = cos 90° = 0
R.H.S. = 4 cos3 A – 3 cos A
= 4 cos3 30° – 3 cos 30°
= `4(sqrt3/2)^3 - 3(sqrt3/2)`
= `(3sqrt3)/2 - (3sqrt3)/2`
= 0
L.H.S. = R.H.S.
APPEARS IN
संबंधित प्रश्न
Prove that `cosA/(1+sinA) + tan A = secA`
If `( cosec theta + cot theta ) =m and ( cosec theta - cot theta ) = n, ` show that mn = 1.
Write the value of tan10° tan 20° tan 70° tan 80° .
Prove the following identity :
`[1/((sec^2θ - cos^2θ)) + 1/((cosec^2θ - sin^2θ))](sin^2θcos^2θ) = (1 - sin^2θcos^2θ)/(2 + sin^2θcos^2θ)`
Without using a trigonometric table, prove that
`(cos 70°)/(sin 20°) + (cos 59°)/(sin 31°) - 8sin^2 30° = 0`.
If cot θ + tan θ = x and sec θ – cos θ = y, then prove that `(x^2y)^(2/3) – (xy^2)^(2/3)` = 1
If tan θ × A = sin θ, then A = ?
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
Prove that `"cot A"/(1 - cot"A") + "tan A"/(1 - tan "A")` = – 1
tan θ × `sqrt(1 - sin^2 θ)` is equal to: