Advertisements
Advertisements
प्रश्न
Prove that:
tan 13x − tan 9x − tan 4x = tan 13x tan 9x tan 4x
Advertisements
उत्तर
\[\text{ We know that }13x = 9x + 4x\]
Therefore,
\[ \tan\left( 13x \right) = \tan\left( 9x + 4x \right)\]
\[ \Rightarrow \tan13x = \frac{\tan9x + \tan4x}{1 - \tan9x \tan4x}\]
\[ \Rightarrow \tan13x - \tan13x \tan9x \tan 4x = \tan9x + \tan4x\]
\[ \Rightarrow \tan13x - \tan9x - \tan4x = \tan13x \tan9x \tan4x \]
Hence proved .
APPEARS IN
संबंधित प्रश्न
Find the value of: sin 75°
Prove the following:
`(sin x - siny)/(cos x + cos y)= tan (x -y)/2`
Prove the following:
`(sin x - sin 3x)/(sin^2 x - cos^2 x) = 2sin x`
Prove the following:
cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1
Prove the following:
cos 4x = 1 – 8sin2 x cos2 x
Prove that: `(cos x + cos y)^2 + (sin x - sin y )^2 = 4 cos^2 (x + y)/2`
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 \[\sin A = \frac{4}{5}\] and \[\cos B = \frac{5}{13}\], where 0 < A, \[B < \frac{\pi}{2}\], find the value of the following:
sin (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:
cos (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).
Evaluate the following:
sin 78° cos 18° − cos 78° sin 18°
Prove that
Prove that:
\[\frac{\tan \left( A + B \right)}{\cot \left( A - B \right)} = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B}\]
If tan A + tan B = a and cot A + cot B = b, prove that cot (A + B) \[\frac{1}{a} - \frac{1}{b}\].
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 α, β are two different values of x lying between 0 and 2π, which satisfy the equation 6 cos x + 8 sin x = 9, find the value of sin (α + β).
If 12 sin x − 9sin2 x attains its maximum value at x = α, then write the value of sin α.
If tan \[\alpha = \frac{1}{1 + 2^{- x}}\] and \[\tan \beta = \frac{1}{1 + 2^{x + 1}}\] then write the value of α + β lying in the interval \[\left( 0, \frac{\pi}{2} \right)\]
If A + B + C = π, then sec A (cos B cos C − sin B sin C) is equal to
If \[\tan A = \frac{a}{a + 1}\text{ and } \tan B = \frac{1}{2a + 1}\]
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
If cos (A − B) \[= \frac{3}{5}\] and tan A tan B = 2, then
If `(sin(x + y))/(sin(x - y)) = (a + b)/(a - b)`, then show that `tanx/tany = a/b` [Hint: Use Componendo and Dividendo].
Find the most general value of θ satisfying the equation tan θ = –1 and cos θ = `1/sqrt(2)`.
If cotθ + tanθ = 2cosecθ, then find the general value of θ.
If cos(θ + Φ) = m cos(θ – Φ), then prove that 1 tan θ = `(1 - m)/(1 + m) cot phi`
[Hint: Express `(cos(theta + Φ))/(cos(theta - Φ)) = m/1` and apply Componendo and Dividendo]
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`]
The value of tan 75° - cot 75° is equal to ______.
The value of tan3A - tan2A - tanA is equal to ______.
The maximum distance of a point on the graph of the function y = `sqrt(3)` sinx + cosx from x-axis is ______.
State whether the statement is True or False? Also give justification.
If cosecx = 1 + cotx then x = 2nπ, 2nπ + `pi/2`
