Advertisements
Advertisements
प्रश्न
If \[\tan A = \frac{a}{a + 1}\text{ and } \tan B = \frac{1}{2a + 1}\]
पर्याय
(a) 0
(b)\[\frac{\pi}{2}\]
(c) \[\frac{\pi}{3}\]
(d) \[\frac{\pi}{4}\]
उत्तर
(d)\[\frac{\pi}{4}\]
\[\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B} \]
\[ = \frac{\frac{a}{a + 1} + \frac{1}{2a + 1}}{1 - \frac{a}{\left( a + 1 \right)(2a + 1)}}\]
\[ = \frac{2 a^2 + a + a + 1}{2 a^2 + 3a + 1 - a}\]
\[ = \frac{2 a^2 + 2a + 1}{2 a^2 + 2a + 1}\]
\[ = 1\]
\[\text{ Therefore }, A + B = \tan^{- 1} (1) = \frac{\pi}{4} . \]
APPEARS IN
संबंधित प्रश्न
Prove that `2 sin^2 pi/6 + cosec^2 (7pi)/6 cos^2 pi/3 = 3/2`
Prove that `cot^2 pi/6 + cosec (5pi)/6 + 3 tan^2 pi/6 = 6`
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`
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 \[\sin A = \frac{12}{13}\text{ and } \sin B = \frac{4}{5}\], where \[\frac{\pi}{2}\] < A < π and 0 < B < \[\frac{\pi}{2}\], find the following:
cos (A + B)
If \[\sin A = \frac{1}{2}, \cos B = \frac{\sqrt{3}}{2}\], where \[\frac{\pi}{2}\] < A < π and 0 < B < \[\frac{\pi}{2}\], find the following:
tan (A - B)
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:
tan (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:
cos2 A + cos2 B − 2 cos A cos B cos (A + B) = sin2 (A + B)
Prove that:
\[\tan\frac{\pi}{12} + \tan\frac{\pi}{6} + \tan\frac{\pi}{12}\tan\frac{\pi}{6} = 1\]
Prove that:
tan 36° + tan 9° + tan 36° tan 9° = 1
Prove that sin2 (n + 1) A − sin2 nA = sin (2n + 1) A sin A.
If tan A = x tan B, prove that
\[\frac{\sin \left( A - B \right)}{\sin \left( A + B \right)} = \frac{x - 1}{x + 1}\]
If tan (A + B) = x and tan (A − B) = y, find the values of tan 2A and tan 2B.
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 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 tan α = x +1, tan β = x − 1, show that 2 cot (α − β) = x2.
Reduce each of the following expressions to the sine and cosine of a single expression:
\[\sqrt{3} \sin x - \cos x\]
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 \[\frac{\cos \left( x - y \right)}{\cos \left( x + y \right)} = \frac{m}{n}\] then write the value of tan x tan y.
tan 3A − tan 2A − tan A =
If cot (α + β) = 0, sin (α + 2β) is equal to
If sin (π cos x) = cos (π sin x), then sin 2x = ______.
If cos (A − B) \[= \frac{3}{5}\] and tan A tan B = 2, then
Match each item given under column C1 to its correct answer given under column C2.
| C1 | C2 |
| (a) `(1 - cosx)/sinx` | (i) `cot^2 x/2` |
| (b) `(1 + cosx)/(1 - cosx)` | (ii) `cot x/2` |
| (c) `(1 + cosx)/sinx` | (iii) `|cos x + sin x|` |
| (d) `sqrt(1 + sin 2x)` | (iv) `tan x/2` |
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 sin(45° + θ) - cos(45° - θ) is ______.
If sinθ + cosθ = 1, then the value of sin2θ 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 cosecx = 1 + cotx then x = 2nπ, 2nπ + `pi/2`
