Advertisements
Advertisements
प्रश्न
If cos A + sin B = m and sin A + cos B = n, prove that 2 sin (A + B) = m2 + n2 − 2.
Advertisements
उत्तर
\[\text{ RHS }= m^2 + n^2 - 2\]
\[ = \left( \cos A + \sin B \right)^2 + \left( \sin A + \cos B \right)^2 - 2\]
\[ = \cos^2 A + \sin^2 B + 2\cos A\sin B + \sin^2 A + \cos {}^2 B + 2\sin A\cos B - 2\]
\[ = 1 + 1 + 2\cos A\sin B + 2\sin A\cos B - 2\]
\[ = 2\left( \cos A\sin B + \sin A \cos B \right)\]
\[ = 2\sin\left( A + B \right)\]
= LHS
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove that: `sin^2 pi/6 + cos^2 pi/3 - tan^2 pi/4 = -1/2`
Prove the following:
`(cos9x - cos5x)/(sin17x - sin 3x) = - (sin2x)/(cos 10x)`
Prove the following:
`(sin x + sin 3x)/(cos x + cos 3x) = tan 2x`
Prove the following:
`(cos 4x + cos 3x + cos 2x)/(sin 4x + sin 3x + sin 2x) = cot 3x`
Prove the following:
cos 4x = 1 – 8sin2 x cos2 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: `(cos x - cosy)^2 + (sin x - sin y)^2 = 4 sin^2 (x - y)/2`
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 \[\sin A = \frac{3}{5}, \cos B = - \frac{12}{13}\], where A and B both lie in second quadrant, find the value of 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 \[\frac{\tan 69^\circ + \tan 66^\circ}{1 - \tan 69^\circ \tan 66^\circ} = - 1\].
If \[\tan A = \frac{m}{m - 1}\text{ and }\tan B = \frac{1}{2m - 1}\], then prove that \[A - B = \frac{\pi}{4}\].
Prove that:
\[\frac{\sin \left( A - B \right)}{\cos A \cos B} + \frac{\sin \left( B - C \right)}{\cos B \cos C} + \frac{\sin \left( C - A \right)}{\cos C \cos A} = 0\]
Prove that:
sin2 B = sin2 A + sin2 (A − B) − 2 sin A cos B sin (A − B)
Prove that:
\[\frac{\tan \left( A + B \right)}{\cot \left( A - B \right)} = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B}\]
Prove that:
tan 13x − tan 9x − tan 4x = tan 13x tan 9x tan 4x
Prove that:
\[\frac{\tan^2 2x - \tan^2 x}{1 - \tan^2 2x \tan^2 x} = \tan 3x \tan x\]
If x lies in the first quadrant and \[\cos x = \frac{8}{17}\], then prove that:
If sin (α + β) = 1 and sin (α − β) \[= \frac{1}{2}\], where 0 ≤ α, \[\beta \leq \frac{\pi}{2}\], then find the values of tan (α + 2β) and tan (2α + β).
Prove that:
If angle \[\theta\] is divided into two parts such that the tangents of one part is \[\lambda\] times the tangent of other, and \[\phi\] is their difference, then show that\[\sin\theta = \frac{\lambda + 1}{\lambda - 1}\sin\phi\]
Find the maximum and minimum values of each of the following trigonometrical expression:
sin x − cos x + 1
Write the maximum and minimum values of 3 cos x + 4 sin x + 5.
If \[\frac{\cos \left( x - y \right)}{\cos \left( x + y \right)} = \frac{m}{n}\] then write the value of tan x tan y.
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
The value of \[\cos^2 \left( \frac{\pi}{6} + x \right) - \sin^2 \left( \frac{\pi}{6} - x \right)\] is
If \[\tan\theta = \frac{1}{2}\] and \[\tan\phi = \frac{1}{3}\], then the value of \[\tan\phi = \frac{1}{3}\] is
If tan (A − B) = 1 and sec (A + B) = \[\frac{2}{\sqrt{3}}\], the smallest positive value of B is
Express the following as the sum or difference of sines and cosines:
2 cos 3x sin 2xa
If sinθ + cosθ = 1, then find the general value of θ.
If sin(θ + α) = a and sin(θ + β) = b, then prove that cos 2(α - β) - 4ab cos(α - β) = 1 - 2a2 - 2b2
[Hint: Express cos(α - β) = cos((θ + α) - (θ + β))]
Find the general solution of the equation `(sqrt(3) - 1) costheta + (sqrt(3) + 1) sin theta` = 2
[Hint: Put `sqrt(3) - 1` = r sinα, `sqrt(3) + 1` = r cosα which gives tanα = `tan(pi/4 - pi/6)` α = `pi/12`]
If sinθ + cosecθ = 2, then sin2θ + cosec2θ is equal to ______.
The value of sin(45° + θ) - cos(45° - θ) is ______.
If sinθ + cosθ = 1, then the value of sin2θ is equal to ______.
