मराठी

If α and β are the solutions of the equation a tan θ + b sec θ = c, then show that tan (α + β) = 2aca2-c2. - Mathematics

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)`.

बेरीज
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)`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Trigonometric Functions - Solved Examples [पृष्ठ ४४]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 3 Trigonometric Functions
Solved Examples | Q 11 | पृष्ठ ४४

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Prove that  `2 sin^2  pi/6 + cosec^2  (7pi)/6 cos^2  pi/3 = 3/2`


Prove the following:

`cos ((3pi)/ 2 + x ) cos(2pi + x) [cot ((3pi)/2 - x) + cot (2pi + x)]= 1`


Prove the following:

cos2 2x – cos2 6x = sin 4x sin 8x


Prove the following:

sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x


Prove the following:

`(sin x - sin 3x)/(sin^2 x - cos^2 x) =  2sin x`


Prove the following:

cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1


Prove that: `(cos x  + cos y)^2 + (sin x - sin y )^2 =  4 cos^2  (x + y)/2`


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).


Evaluate the following:
sin 36° cos 9° + cos 36° sin 9°


Evaluate the following:
 cos 80° cos 20° + sin 80° sin 20°


Prove that

\[\frac{\cos 11^\circ + \sin 11^\circ}{\cos 11^\circ - \sin 11^\circ} = \tan 56^\circ\]

Prove that

\[\frac{\cos 9^\circ + \sin 9^\circ}{\cos 9^\circ - \sin 9^\circ} = \tan 54^\circ\]

Prove that:

\[\sin\left( \frac{3\pi}{8} - 5 \right)\cos\left( \frac{\pi}{8} + 5 \right) + \cos\left( \frac{3\pi}{8} - 5 \right)\sin\left( \frac{\pi}{8} + 5 \right) = 1\]

 


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:

\[\frac{\sin \left( A - B \right)}{\sin A \sin B} + \frac{\sin \left( B - C \right)}{\sin B \sin C} + \frac{\sin \left( C - A \right)}{\sin C \sin A} = 0\]

 


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 sin α + sin β = a and cos α + cos β = b, show that

\[\sin \left( \alpha + \beta \right) = \frac{2ab}{a^2 + b^2}\]

 


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 sin α sin β − cos α cos β + 1 = 0, prove that 1 + cot α tan β = 0.


If tan α = x +1, tan β = x − 1, show that 2 cot (α − β) = x2.


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. 


The value of \[\sin^2 \frac{5\pi}{12} - \sin^2 \frac{\pi}{12}\] 


tan 20° + tan 40° + \[\sqrt{3}\] tan 20° tan 40° is equal to 


If 3 sin x + 4 cos x = 5, then 4 sin x − 3 cos x =


If A + B + C = π, then \[\frac{\tan A + \tan B + \tan C}{\tan A \tan B \tan C}\] is equal to

 

If cot (α + β) = 0, sin (α + 2β) is equal to


If tan (A − B) = 1 and sec (A + B) = \[\frac{2}{\sqrt{3}}\], the smallest positive value of B is

 

Express the following as the sum or difference of sines and cosines:
2 sin 4x sin 3x


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 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 tanA = `1/2`, tanB = `1/3`, then tan(2A + B) is equal to ______.


If sinx + cosx = a, then sin6x + cos6x = ______.


Given x > 0, the values of f(x) = `-3cos sqrt(3 + x + x^2)` lie in the interval ______.


State whether the statement is True or False? Also give justification.

If tan(π cosθ) = cot(π sinθ), then `cos(theta - pi/4) = +- 1/(2sqrt(2))`.


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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×