Advertisements
Advertisements
Question
Prove the following: `cos (pi/4 xx x) cos (pi/4 - y) - sin (pi/4 - x)sin (pi/4 - y) = sin (x + y)`
Solution
L.H.S. `cos (pi/4 xx x) cos (pi/4 - y) - sin (pi/4 - x)sin (pi/4 - y)`
= cos `[(pi/4 - x + pi/4 - y)]`
[∵ cos A cos B - sin A sin B = cos (A + B)]
= cos `(pi/2 - (x +y))`
= sin (x + y) = R.H.S. `(∵ cos (pi/2 - θ) = sin θ)`
APPEARS IN
RELATED QUESTIONS
Prove the following: `(tan(pi/4 + x))/(tan(pi/4 - x)) = ((1+ tan x)/(1- tan x))^2`
Prove the following:
sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
Prove the following:
`cos ((3pi)/4 + x) - cos((3pi)/4 - x) = -sqrt2 sin x`
Prove that: `((sin 7x + sin 5x) + (sin 9x + sin 3x))/((cos 7x + cos 5x) + (cos 9x + cos 3x)) = tan 6x`
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 \[\tan A = \frac{3}{4}, \cos B = \frac{9}{41}\], where π < A < \[\frac{3\pi}{2}\] and 0 < B <\[\frac{\pi}{2}\], find tan (A + B).
Evaluate the following:
sin 36° cos 9° + cos 36° sin 9°
If \[\cos A = - \frac{12}{13}\text{ and }\cot B = \frac{24}{7}\], where A lies in the second quadrant and B in the third quadrant, find the values of the following:
sin (A + B)
If \[\cos A = - \frac{12}{13}\text{ and }\cot B = \frac{24}{7}\], where A lies in the second quadrant and B in the third quadrant, find the values of the following:
cos (A + B)
Prove that
\[\frac{\tan A + \tan B}{\tan A - \tan B} = \frac{\sin \left( A + B \right)}{\sin \left( A - B \right)}\]
Prove that:
Prove that \[\frac{\tan 69^\circ + \tan 66^\circ}{1 - \tan 69^\circ \tan 66^\circ} = - 1\].
If \[\tan A = \frac{5}{6}\text{ and }\tan B = \frac{1}{11}\], prove that \[A + B = \frac{\pi}{4}\].
Prove that:
\[\cos^2 45^\circ - \sin^2 15^\circ = \frac{\sqrt{3}}{4}\]
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)
If tan x + \[\tan \left( x + \frac{\pi}{3} \right) + \tan \left( x + \frac{2\pi}{3} \right) = 3\], then prove that \[\frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x} = 1\].
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).
If sin α sin β − cos α cos β + 1 = 0, prove that 1 + cot α tan β = 0.
If \[\tan\theta = \frac{\sin\alpha - \cos\alpha}{\sin\alpha + \cos\alpha}\] , then show that \[\sin\alpha + \cos\alpha = \sqrt{2}\cos\theta\].
If α + β − γ = π and sin2 α +sin2 β − sin2 γ = λ sin α sin β cos γ, then write the value of λ.
If x cos θ = y cos \[\left( \theta + \frac{2\pi}{3} \right) = z \cos \left( \theta + \frac{4\pi}{3} \right)\]then write the value of \[\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\]
If tan (A + B) = p and tan (A − B) = q, then write the value of tan 2B.
The value of \[\sin^2 \frac{5\pi}{12} - \sin^2 \frac{\pi}{12}\]
tan 3A − tan 2A − tan A =
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 (36° − A) cos (36° + A) + cos (54° + A) cos (54° − A) is
If tan 69° + tan 66° − tan 69° tan 66° = 2k, then k =
If \[\tan\alpha = \frac{x}{x + 1}\] and \[\tan\alpha = \frac{x}{x + 1}\], then \[\alpha + \beta\] is equal to
If 3 tan (θ – 15°) = tan (θ + 15°), 0° < θ < 90°, then θ = ______.
If tanθ = `(sinalpha - cosalpha)/(sinalpha + cosalpha)`, then show that sinα + cosα = `sqrt(2)` cosθ.
[Hint: Express tanθ = `tan (alpha - pi/4) theta = alpha - pi/4`]
If sinθ + cosθ = 1, then find the general value of θ.
If tanA = `1/2`, tanB = `1/3`, then tan(2A + B) is equal to ______.
If tanα = `1/7`, tanβ = `1/3`, then cos2α is equal to ______.
3(sinx – cosx)4 + 6(sinx + cosx)2 + 4(sin6x + cos6x) = ______.
State whether the statement is True or False? Also give justification.
If tanA = `(1 - cos B)/sinB`, then tan2A = tanB
State whether the statement is True or False? Also give justification.
If cosecx = 1 + cotx then x = 2nπ, 2nπ + `pi/2`
