Advertisements
Advertisements
प्रश्न
Prove the following:
cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1
Advertisements
उत्तर
cot 3x = cot(2x + x) = `(cot 2x cot x-1)/(cot x + cot 2x)` `[∵ cot (A + B) = (cot A cot B - 1)/(cot A + cot B)]`
⇒ cot 3x (cot x + cot 2x) = cot 2x cot x - 1
⇒ cot 3x cot x + cot 3x cot 2x = cot 2x cot x - 1
⇒ cot x cot 2x - cot 2x cot 3x - cot 3x cot x = 1
APPEARS IN
संबंधित प्रश्न
Find the value of: sin 75°
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:
sin2 6x – sin2 4x = sin 2x sin 10x
Prove the following:
`(sin 5x + sin 3x)/(cos 5x + cos 3x) = tan 4x`
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 \[\sin A = \frac{3}{5}, \cos B = - \frac{12}{13}\], where A and B both lie in second quadrant, find the value of sin (A + B).
If \[\sin A = \frac{1}{2}, \cos B = \frac{12}{13}\], where \[\frac{\pi}{2}\]< A < π and \[\frac{3\pi}{2}\] < B < 2π, find tan (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)
Evaluate the following:
sin 78° cos 18° − cos 78° sin 18°
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)
Prove that:
\[\frac{7\pi}{12} + \cos\frac{\pi}{12} = \sin\frac{5\pi}{12} - \sin\frac{\pi}{12}\]
Prove that
Prove that:
Prove that:
Prove that \[\frac{\tan 69^\circ + \tan 66^\circ}{1 - \tan 69^\circ \tan 66^\circ} = - 1\].
Prove that:
cos2 A + cos2 B − 2 cos A cos B cos (A + B) = sin2 (A + B)
Prove that:
tan 36° + tan 9° + tan 36° tan 9° = 1
If tan A + tan B = a and cot A + cot B = b, prove that cot (A + B) \[\frac{1}{a} - \frac{1}{b}\].
If x lies in the first quadrant and \[\cos x = \frac{8}{17}\], then prove that:
Prove that:
Show that sin 100° − sin 10° is positive.
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 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 \[\left( 0, \frac{\pi}{2} \right)\]
The value of \[\sin^2 \frac{5\pi}{12} - \sin^2 \frac{\pi}{12}\]
tan 3A − tan 2A − tan A =
If A − B = π/4, then (1 + tan A) (1 − tan B) is equal to
If α and β are the solutions of the equation a tan θ + b sec θ = c, then show that tan (α + β) = `(2ac)/(a^2 - c^2)`.
If sinθ + cosθ = 1, 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]
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 tanθ = `a/b`, then bcos2θ + asin2θ is equal to ______.
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
