English

If Cos X = Cos α + Cos β 1 + Cos α Cos β , Prove that Tan X 2 = ± Tan α 2 Tan β 2 - Mathematics

Advertisements
Advertisements

Question

If \[\cos x = \frac{\cos \alpha + \cos \beta}{1 + \cos \alpha \cos \beta}\] , prove that \[\tan\frac{x}{2} = \pm \tan\frac{\alpha}{2}\tan\frac{\beta}{2}\]

 
Numerical
Advertisements

Solution

Given: \[\cos x = \frac{\cos \alpha + \cos \beta}{1 + \cos \alpha \cos \beta}\]        ...(1)

\[\Rightarrow \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} = \frac{cos\alpha + cos\beta}{1 + cos\alpha \times cos\beta} \left[ \because \text{ cos } x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right]\]
\[\text{ By componendo and dividendo, we get } \]
\[\frac{\left( 1 - \tan^2 \frac{x}{2} \right) + \left( 1 + \tan^2 \frac{x}{2} \right)}{\left( 1 - \tan^2 \frac{x}{2} \right) - \left( 1 + \tan^2 \frac{x}{2} \right)} = \frac{\left( 1 + cos\alpha \times cos\beta + cos\alpha + cos\beta \right)}{- \left( 1 + cos\alpha cos\beta - cos\alpha - cos\beta \right)}\]
\[ \Rightarrow \frac{2}{2 \tan^2 \frac{x}{2}} = \frac{\left( 1 + cos\alpha \right)\left( 1 + cos\beta \right)}{\left( 1 - cos\alpha \right)\left( 1 - cos\beta \right)}\]

\[\Rightarrow \tan^2 \frac{x}{2} = \frac{\left( 1 - cos\alpha \right)\left( 1 - cos\beta \right)}{\left( 1 + cos\alpha \right)\left( 1 + cos\beta \right)}\]
\[ \Rightarrow \tan^2 \frac{x}{2} = \frac{2 \sin^2 \frac{\alpha}{2} \times 2 \sin^2 \frac{\beta}{2}}{2 \cos^2 \frac{\alpha}{2} \times 2 \cos^2 \frac{\beta}{2}}\]
\[ \Rightarrow \tan^2 \frac{x}{2} = \tan^2 \frac{\alpha}{2} \times \tan^2 \frac{\beta}{2}\]
\[ \Rightarrow \tan\frac{x}{2} = \pm \tan\frac{\alpha}{2} \times \tan\frac{\beta}{2}\]
\[\text{ Hence proved }  .\]

shaalaa.com
Values of Trigonometric Functions at Multiples and Submultiples of an Angle
  Is there an error in this question or solution?
Chapter 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.1 [Page 29]

APPEARS IN

RD Sharma Mathematics [English] Class 11
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 40 | Page 29

RELATED QUESTIONS

Prove that:  \[\frac{1 - \cos 2x + \sin 2x}{1 + \cos 2x + \sin 2x} = \tan x\]

 

Prove that:  \[\frac{\cos 2 x}{1 + \sin 2 x} = \tan \left( \frac{\pi}{4} - x \right)\]

 

Prove that: \[\sin^2 \frac{\pi}{8} + \sin^2 \frac{3\pi}{8} + \sin^2 \frac{5\pi}{8} + \sin^2 \frac{7\pi}{8} = 2\]


Prove that: \[1 + \cos^2 2x = 2 \left( \cos^4 x + \sin^4 x \right)\]

 

Prove that: \[\left( \sin 3x + \sin x \right) \sin x + \left( \cos 3x - \cos x \right) \cos x = 0\]


Prove that: \[\sin 4x = 4 \sin x \cos^3 x - 4 \cos x \sin^3 x\]

 

Show that: \[2 \left( \sin^6 x + \cos^6 x \right) - 3 \left( \sin^4 x + \cos^4 x \right) + 1 = 0\]

 

If  \[\sin x = \frac{\sqrt{5}}{3}\] and x lies in IInd quadrant, find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2} \text{ and }  \tan \frac{x}{2}\] . 

 

 


Prove that: \[\cos\frac{\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5} = \frac{- 1}{16}\]

 

Prove that: \[\cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} = \frac{1}{64}\]

 

If  \[\sec \left( x + \alpha \right) + \sec \left( x - \alpha \right) = 2 \sec x\] , prove that \[\cos x = \pm \sqrt{2} \cos\frac{\alpha}{2}\]

 

If  \[\sin \alpha = \frac{4}{5} \text{ and }  \cos \beta = \frac{5}{13}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \frac{8}{\sqrt{65}}\]

 

Prove that `tan x + tan (π/3 + x) - tan(π/3 - x) = 3tan 3x`


Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]

  

Prove that: \[\cos 6° \cos 42°   \cos 66°    \cos 78° = \frac{1}{16}\]

 

If \[\tan\frac{x}{2} = \frac{m}{n}\] , then write the value of m sin x + n cos x.

 

 


If \[\frac{\pi}{2} < x < \pi,\] the write the value of \[\sqrt{2 + \sqrt{2 + 2 \cos 2x}}\] in the simplest form.

 
 

If  \[\text{ sin } x + \text{ cos } x = a\], then find the value of

\[\sin^6 x + \cos^6 x\] .
 

 


If  \[\text{ sin } x + \text{ cos } x = a\], find the value of \[\left|\text { sin } x - \text{ cos } x \right|\] .

 

 


\[\frac{\sec 8A - 1}{\sec 4A - 1} =\]

 


If \[\cos 2x + 2 \cos x = 1\]  then, \[\left( 2 - \cos^2 x \right) \sin^2 x\]  is equal to 

 
 

For all real values of x, \[\cot x - 2 \cot 2x\] is equal to 

 

If in a  \[∆ ABC, \tan A + \tan B + \tan C = 0\], then

\[\cot A \cot B \cot C =\]
 

 


If \[\tan \alpha = \frac{1 - \cos \beta}{\sin \beta}\] , then

 

If \[\sin \alpha + \sin \beta = a \text{ and }  \cos \alpha - \cos \beta = b \text{ then }  \tan \frac{\alpha - \beta}{2} =\]

 


The value of \[\left( \cot \frac{x}{2} - \tan \frac{x}{2} \right)^2 \left( 1 - 2 \tan x \cot 2 x \right)\] is 

 

If \[A = 2 \sin^2 x - \cos 2x\] , then A lies in the interval


The value of \[\frac{2\left( \sin 2x + 2 \cos^2 x - 1 \right)}{\cos x - \sin x - \cos 3x + \sin 3x}\] is 

 

\[2 \left( 1 - 2 \sin^2 7x \right) \sin 3x\]  is equal to


If  \[\tan \frac{x}{2} = \frac{\sqrt{1 - e}}{1 + e} \tan \frac{\alpha}{2}\] , then \[\cos \alpha =\]


If  \[\left( 2^n + 1 \right) x = \pi,\] then \[2^n \cos x \cos 2x \cos 2^2 x . . . \cos 2^{n - 1} x = 1\]

 


The value of \[\tan x \tan \left( \frac{\pi}{3} - x \right) \tan \left( \frac{\pi}{3} + x \right)\] is

 

The value of `cos^2 48^@ - sin^2 12^@` is ______.


If tanθ + sinθ = m and tanθ – sinθ = n, then prove that m2 – n2 = 4sinθ tanθ 
[Hint: m + n = 2tanθ, m – n = 2sinθ, then use m2 – n2 = (m + n)(m – n)]


Find the value of the expression `cos^4  pi/8 + cos^4  (3pi)/8 + cos^4  (5pi)/8 + cos^4  (7pi)/8`

[Hint: Simplify the expression to `2(cos^4  pi/8 + cos^4  (3pi)/8) = 2[(cos^2  pi/8 + cos^2  (3pi)/8)^2 - 2cos^2  pi/8 cos^2  (3pi)/8]`


The value of cos12° + cos84° + cos156° + cos132° is ______.


The value of `sin  pi/18 + sin  pi/9 + sin  (2pi)/9 + sin  (5pi)/18` is given by ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×