Advertisements
Advertisements
प्रश्न
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)`.
Advertisements
उत्तर
Given that atanθ + bsecθ = c or asinθ + b = c cos θ
Using the identities,
sin θ = `(2tan theta/2)/(1 + tan^2 theta/2)` and cos θ = `(1 - tan^2 theta/2)/(1 + tan^2 theta/2)`
We have, `(a(2tan theta/2))/(1 + tan^2 theta/2) + b = (c(1 - tan^2 theta/2))/(1 + tan^2 theta/2)`
or `(b + c) tan^2 theta/2 + 2a tan theta/2 + b - c` = 0
The above equation is quadratic in `tan theta/2` and hence `tan alpha/2` and `tan beta/2` are the roots of this equation.
Therefore, `tan alpha/2 + tan beta/2 = (-2a)/(b + c)` and `tan alpha/2 tan beta/2 - (b - c)/(b + c)`
Using the identity `tan(alpha/2 + beta/2) = (tan alpha/2 + tan beta/2)/(1 - tan alpha/2 tan beta/2)`
We have, `tan(alpha/2 + beta/2) = ((-2a)/(b + c))/(1 - (b - c)/(b + c))`
= `(-2a)/(2c) = (-a)/c` .....(1)
Again, using another identity
`tan 2 (alpha + beta)/2 = (2tan (alpha + beta)/2)/(1 - tan^2 (alpha + beta)/2)`
We have tan (α + β) = `(2(- a/c))/(1 - a^2/c^2)`
= `(2ac)/(a^2 - c^2)` ......[From (1)]
Alternatively, given that a tanθ + b secθ = c
⇒ (a tanθ – c)2 = b2 (1 + tan2θ)
⇒ a2 tan2θ – 2ac tanθ + c2 = b2 + b2 tan2θ
⇒ (a2 – b2) tan2θ – 2ac tanθ + c2 – b2 = 0 ......(1)
Since α and β are the roots of the equation (1)
So tanα + tanβ = `(2ac)/(a^2 - b^2)`
And tanα tanβ = `(c^2 - b^2)/(a^2 - b^2)`
Therefore, tan (α + β) = `(tan alpha + tan beta)/(1 - tan alpha tan beta)`
= `((2ac)/(a^2 - b^2))/((c^2 - b^2)/(a^2 - b^2))`
= `(2ac)/(a^2 - c^2)`
APPEARS IN
संबंधित प्रश्न
Prove that `cot^2 pi/6 + cosec (5pi)/6 + 3 tan^2 pi/6 = 6`
Prove the following:
sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x
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:
`(cos 4x + cos 3x + cos 2x)/(sin 4x + sin 3x + sin 2x) = cot 3x`
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:
cos 47° cos 13° − sin 47° sin 13°
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
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:
\[\frac{\sin \left( A - B \right)}{\cos A \cos B} + \frac{\sin \left( B - C \right)}{\cos B \cos C} + \frac{\sin \left( C - A \right)}{\cos C \cos A} = 0\]
Prove that:
\[\frac{\tan \left( A + B \right)}{\cot \left( A - B \right)} = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B}\]
Prove that:
tan 8x − tan 6x − tan 2x = tan 8x tan 6x tan 2x
Prove that sin2 (n + 1) A − sin2 nA = sin (2n + 1) A sin A.
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 tan α = x +1, tan β = x − 1, show that 2 cot (α − β) = x2.
Find the maximum and minimum values of each of the following trigonometrical expression:
12 cos x + 5 sin x + 4
Find the maximum and minimum values of each of the following trigonometrical expression:
\[5 \cos x + 3 \sin \left( \frac{\pi}{6} - x \right) + 4\]
Prove that \[\left( 2\sqrt{3} + 3 \right) \sin x + 2\sqrt{3} \cos x\] lies between \[- \left( 2\sqrt{3} + \sqrt{15} \right) \text{ and } \left( 2\sqrt{3} + \sqrt{15} \right)\]
Write the maximum and minimum values of 3 cos x + 4 sin x + 5.
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 \[\sin^2 \frac{5\pi}{12} - \sin^2 \frac{\pi}{12}\]
If \[\tan A = \frac{a}{a + 1}\text{ and } \tan B = \frac{1}{2a + 1}\]
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
If tan θ1 tan θ2 = k, then \[\frac{\cos \left( \theta_1 - \theta_2 \right)}{\cos \left( \theta_1 + \theta_2 \right)} =\]
Express the following as the sum or difference of sines and cosines:
2 sin 3x cos x
Express the following as the sum or difference of sines and cosines:
2 sin 4x sin 3x
If `(sin(x + y))/(sin(x - y)) = (a + b)/(a - b)`, then show that `tanx/tany = a/b` [Hint: Use Componendo and Dividendo].
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 ______.
If sinx + cosx = a, then |sinx – cosx| = ______.
Given x > 0, the values of f(x) = `-3cos sqrt(3 + x + x^2)` lie in the interval ______.
In the following match each item given under the column C1 to its correct answer given under the column C2:
| Column A | Column B |
| (a) sin(x + y) sin(x – y) | (i) cos2x – sin2y |
| (b) cos (x + y) cos (x – y) | (ii) `(1 - tan theta)/(1 + tan theta)` |
| (c) `cot(pi/4 + theta)` | (iii) `(1 + tan theta)/(1 - tan theta)` |
| (d) `tan(pi/4 + theta)` | (iv) sin2x – sin2y |
