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 (3pi)/4 + 2 cos^2 pi/4 + 2 sec^2 pi/3 = 10`
Prove the following:
`cos ((3pi)/4 + x) - cos((3pi)/4 - x) = -sqrt2 sin x`
Prove the following:
`(sin x - sin 3x)/(sin^2 x - cos^2 x) = 2sin x`
Prove that: sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 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:
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:
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)
Prove that
Prove that:
Prove that:
sin2 (n + 1) A − sin2 nA = sin (2n + 1) A sin 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:
sin2 B = sin2 A + sin2 (A − B) − 2 sin A cos B sin (A − B)
Prove that:
cos2 A + cos2 B − 2 cos A cos B cos (A + B) = sin2 (A + 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 + tan B = a and cot A + cot B = b, prove that cot (A + B) \[\frac{1}{a} - \frac{1}{b}\].
Prove that:
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:
sin x − cos x + 1
Reduce each of the following expressions to the sine and cosine of a single expression:
24 cos x + 7 sin x
Write the interval in which the value of 5 cos x + 3 cos \[\left( x + \frac{\pi}{3} \right) + 3\] lies.
If tan (A + B) = p and tan (A − B) = q, then write the value of tan 2B.
If \[\frac{\cos \left( x - y \right)}{\cos \left( x + y \right)} = \frac{m}{n}\] then write the value of tan x tan y.
If 3 sin x + 4 cos x = 5, then 4 sin x − 3 cos x =
If cos (A − B) \[= \frac{3}{5}\] and tan A tan B = 2, then
If tan 69° + tan 66° − tan 69° tan 66° = 2k, then k =
If \[\tan\alpha = \frac{x}{x + 1}\] and \[\tan\alpha = \frac{x}{x + 1}\], then \[\alpha + \beta\] is equal to
Show that 2 sin2β + 4 cos (α + β) sin α sin β + cos 2(α + β) = cos 2α
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 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]
The value of tan 75° - cot 75° is equal to ______.
If tanα = `m/(m + 1)`, tanβ = `1/(2m + 1)`, then α + β is equal to ______.
If tanA = `1/2`, tanB = `1/3`, then tan(2A + B) is equal to ______.
If sinθ + cosθ = 1, then the value of sin2θ is equal to ______.
