Advertisements
Advertisements
प्रश्न
Prove that:
Advertisements
उत्तर
\[\frac{\pi}{3} = 60^\circ, \frac{\pi}{6} = 30^\circ\]
\[\text{ LHS }= \sin\left( 60^\circ - x \right) \cos\left( 30^\circ + x \right) + \cos\left( 60^\circ - x \right) \sin\left( 30^\circ + x \right)\]
\[ = \sin\left[ \left( 60^\circ - x \right) + \left( 30^\circ + x \right) \right] (\text{ Using the formula }\sin A \cos B + \cos A \sin B = \sin\left( A + B \right) \]
\[\text{ and taking }A = 60^\circ - x\text{ and }B = 30^\circ + x \]
\[ = \sin90^\circ\]
\[ = 1\]
= RHS
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove the following:
sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
Prove the following:
cos2 2x – cos2 6x = sin 4x sin 8x
Prove the following:
`(cos 4x + cos 3x + cos 2x)/(sin 4x + sin 3x + sin 2x) = cot 3x`
Prove the following:
`tan 4x = (4tan x(1 - tan^2 x))/(1 - 6tan^2 x + tan^4 x)`
Prove the following:
cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1
Prove that: `(cos x + cos y)^2 + (sin x - sin y )^2 = 4 cos^2 (x + y)/2`
Prove that: sin 3x + sin 2x – sin x = 4sin x `cos x/2 cos (3x)/2`
If \[\sin A = \frac{4}{5}\] and \[\cos B = \frac{5}{13}\], where 0 < A, \[B < \frac{\pi}{2}\], find the value of the following:
cos (A + B)
If \[\sin A = \frac{12}{13}\text{ and } \sin B = \frac{4}{5}\], where \[\frac{\pi}{2}\] < A < π and 0 < B < \[\frac{\pi}{2}\], find the following:
sin (A + B)
If \[\cos A = - \frac{24}{25}\text{ and }\cos B = \frac{3}{5}\], where π < A < \[\frac{3\pi}{2}\text{ and }\frac{3\pi}{2}\]< B < 2π, find the following:
sin (A + B)
If \[\sin A = \frac{1}{2}, \cos B = \frac{\sqrt{3}}{2}\], where \[\frac{\pi}{2}\] < A < π and 0 < B < \[\frac{\pi}{2}\], find the following:
tan (A + B)
Prove that
Prove that \[\frac{\tan 69^\circ + \tan 66^\circ}{1 - \tan 69^\circ \tan 66^\circ} = - 1\].
Prove that:
Prove that:
cos2 A + cos2 B − 2 cos A cos B cos (A + B) = sin2 (A + B)
Prove that:
tan 13x − tan 9x − tan 4x = tan 13x tan 9x tan 4x
Prove that sin2 (n + 1) A − sin2 nA = sin (2n + 1) A sin A.
If sin α sin β − cos α cos β + 1 = 0, prove that 1 + cot α tan β = 0.
Find the maximum and minimum values of each of the following trigonometrical expression:
12 sin x − 5 cos x
Write the maximum and minimum values of 3 cos x + 4 sin x + 5.
Write the maximum value of 12 sin x − 9 sin2 x.
If \[\cos P = \frac{1}{7}\text{ and }\cos Q = \frac{13}{14}\], where P and Q both are acute angles. Then, the value of P − Q is
If cot (α + β) = 0, sin (α + 2β) is equal to
If \[\tan\theta = \frac{1}{2}\] and \[\tan\phi = \frac{1}{3}\], then the value of \[\tan\phi = \frac{1}{3}\] is
If cos (A − B) \[= \frac{3}{5}\] and tan A tan B = 2, then
If tan 69° + tan 66° − tan 69° tan 66° = 2k, then k =
Express the following as the sum or difference of sines and cosines:
2 sin 3x cos x
Express the following as the sum or difference of sines and cosines:
2 cos 3x sin 2xa
Express the following as the sum or difference of sines and cosines:
2 cos 7x cos 3x
If angle θ is divided into two parts such that the tangent of one part is k times the tangent of other, and Φ is their difference, then show that sin θ = `(k + 1)/(k - 1)` sin Φ
If cotθ + tanθ = 2cosecθ, then find the general value of θ.
If sinθ + cosecθ = 2, then sin2θ + cosec2θ is equal to ______.
If tanα = `m/(m + 1)`, tanβ = `1/(2m + 1)`, then α + β is equal to ______.
The value of tan3A - tan2A - tanA is equal to ______.
The value of `cot(pi/4 + theta)cot(pi/4 - theta)` is ______.
If sinx + cosx = a, then sin6x + cos6x = ______.
State whether the statement is True or False? Also give justification.
If tanA = `(1 - cos B)/sinB`, then tan2A = tanB
In the following match each item given under the column C1 to its correct answer given under the column C2:
| Column A | Column B |
| (a) sin(x + y) sin(x – y) | (i) cos2x – sin2y |
| (b) cos (x + y) cos (x – y) | (ii) `(1 - tan theta)/(1 + tan theta)` |
| (c) `cot(pi/4 + theta)` | (iii) `(1 + tan theta)/(1 - tan theta)` |
| (d) `tan(pi/4 + theta)` | (iv) sin2x – sin2y |
