Advertisements
Advertisements
Question
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
Solution
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
RELATED QUESTIONS
Prove that `2 sin^2 pi/6 + cosec^2 (7pi)/6 cos^2 pi/3 = 3/2`
Prove the following: `cos (pi/4 xx x) cos (pi/4 - y) - sin (pi/4 - x)sin (pi/4 - y) = sin (x + y)`
Prove the following: `(tan(pi/4 + x))/(tan(pi/4 - x)) = ((1+ tan x)/(1- tan x))^2`
Prove the following:
`cos ((3pi)/ 2 + x ) cos(2pi + x) [cot ((3pi)/2 - x) + cot (2pi + x)]= 1`
Prove the following:
sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x
Prove that: `((sin 7x + sin 5x) + (sin 9x + sin 3x))/((cos 7x + cos 5x) + (cos 9x + cos 3x)) = tan 6x`
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 \[\cos A = - \frac{24}{25}\text{ and }\cos B = \frac{3}{5}\], where π < A < \[\frac{3\pi}{2}\text{ and }\frac{3\pi}{2}\]< B < 2π, find the following:
cos (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:
cos 80° cos 20° + sin 80° sin 20°
Prove that
If \[\tan A = \frac{m}{m - 1}\text{ and }\tan B = \frac{1}{2m - 1}\], then prove that \[A - B = \frac{\pi}{4}\].
Prove that: \[\frac{\sin \left( A + B \right) + \sin \left( A - B \right)}{\cos \left( A + B \right) + \cos \left( A - B \right)} = \tan A\]
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:
tan 13x − tan 9x − tan 4x = tan 13x tan 9x tan 4x
If x lies in the first quadrant and \[\cos x = \frac{8}{17}\], then prove that:
If sin α + sin β = a and cos α + cos β = b, show that
Prove that:
\[\frac{1}{\sin \left( x - a \right) \sin \left( x - b \right)} = \frac{\cot \left( x - a \right) - \cot \left( x - b \right)}{\sin \left( a - b \right)}\]
If tan α = x +1, tan β = x − 1, show that 2 cot (α − β) = x2.
If \[\tan\theta = \frac{\sin\alpha - \cos\alpha}{\sin\alpha + \cos\alpha}\] , then show that \[\sin\alpha + \cos\alpha = \sqrt{2}\cos\theta\].
Find the maximum and minimum values of each of the following trigonometrical expression:
\[5 \cos x + 3 \sin \left( \frac{\pi}{6} - x \right) + 4\]
Find the maximum and minimum values of each of the following trigonometrical expression:
sin x − cos x + 1
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 12 sin x − 9sin2 x attains its maximum value at x = α, then write the value of sin α.
Write the interval in which the value of 5 cos x + 3 cos \[\left( x + \frac{\pi}{3} \right) + 3\] lies.
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 in ∆ABC, tan A + tan B + tan C = 6, then cot A cot B cot C =
If cot (α + β) = 0, sin (α + 2β) is equal to
If sin (π cos x) = cos (π sin x), then sin 2x = ______.
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
Express the following as the sum or difference of sines and cosines:
2 cos 7x cos 3x
If sin(θ + α) = a and sin(θ + β) = b, then prove that cos 2(α - β) - 4ab cos(α - β) = 1 - 2a2 - 2b2
[Hint: Express cos(α - β) = cos((θ + α) - (θ + β))]
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 tan θ = 3 and θ lies in third quadrant, then the value of sin θ ______.
If α + β = `pi/4`, then the value of (1 + tan α)(1 + tan β) is ______.
The maximum distance of a point on the graph of the function y = `sqrt(3)` sinx + cosx from x-axis is ______.
