हिंदी

In a ∆Abc, Prove That: Tan a + B 2 = Cot C 2 - Mathematics

Advertisements
Advertisements

प्रश्न

In a ∆ABC, prove that:

\[\tan\frac{A + B}{2} = \cot\frac{C}{2}\]
Advertisements

उत्तर

 In ∆ ABC: 
\[A + B + C = \pi\]
\[ \Rightarrow A + B = \pi - C\]
\[ \Rightarrow \frac{A + B}{2} = \frac{\pi - C}{2}\]
\[ \Rightarrow \frac{A + B}{2} = \frac{\pi}{2} - \frac{C}{2}\]
\[\text{ Now, LHS }= \tan\left( \frac{A + B}{2} \right) \]
\[ = \tan\left( \frac{\pi}{2} - \frac{C}{2} \right) \]
\[ = \cot\left( \frac{C}{2} \right) \left[ \because \tan\left( \frac{\pi}{2} - \theta \right) = \cot \theta \right] \]
 = RHS
Hence proved. 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Trigonometric Functions - Exercise 5.3 [पृष्ठ ४०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 5 Trigonometric Functions
Exercise 5.3 | Q 6.3 | पृष्ठ ४०

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

If \[\cot x \left( 1 + \sin x \right) = 4 m \text{ and }\cot x \left( 1 - \sin x \right) = 4 n,\] \[\left( m^2 + n^2 \right)^2 = mn\]


If \[T_n = \sin^n x + \cos^n x\], prove that  \[2 T_6 - 3 T_4 + 1 = 0\]


Prove that cos 570° sin 510° + sin (−330°) cos (−390°) = 0

Prove that:
\[\frac{\cos (2\pi + x) cosec (2\pi + x) \tan (\pi/2 + x)}{\sec(\pi/2 + x)\cos x \cot(\pi + x)} = 1\]

 


Prove that

\[\frac{\sin(180^\circ + x) \cos(90^\circ + x) \tan(270^\circ - x) \cot(360^\circ - x)}{\sin(360^\circ - x) \cos(360^\circ + x) cosec( - x) \sin(270^\circ + x)} = 1\]

 


Prove that

\[\left\{ 1 + \cot x - \sec\left( \frac{\pi}{2} + x \right) \right\}\left\{ 1 + \cot x + \sec\left( \frac{\pi}{2} + x \right) \right\} = 2\cot x\]

 


Prove that:
\[\sin \frac{13\pi}{3}\sin\frac{2\pi}{3} + \cos\frac{4\pi}{3}\sin\frac{13\pi}{6} = \frac{1}{2}\]


If \[0 < x < \frac{\pi}{2}\], and if \[\frac{y + 1}{1 - y} = \sqrt{\frac{1 + \sin x}{1 - \sin x}}\], then y is equal to


If tan \[x = - \frac{1}{\sqrt{5}}\] and θ lies in the IV quadrant, then the value of cos x is

 

If \[cosec x - \cot x = \frac{1}{2}, 0 < x < \frac{\pi}{2},\]

 

\[\sec^2 x = \frac{4xy}{(x + y )^2}\] is true if and only if

 


If \[cosec x + \cot x = \frac{11}{2}\], then tan x =

 


If \[f\left( x \right) = \cos^2 x + \sec^2 x\], then


Find the general solution of the following equation:

\[\sin 9x = \sin x\]

Find the general solution of the following equation:

\[\sin 2x = \cos 3x\]

Find the general solution of the following equation:

\[\tan 3x = \cot x\]

Find the general solution of the following equation:

\[\sin 2x + \cos x = 0\]

Solve the following equation:

\[\cos x \cos 2x \cos 3x = \frac{1}{4}\]

Solve the following equation:

\[\cos x + \sin x = \cos 2x + \sin 2x\]

Solve the following equation:
3 – 2 cos x – 4 sin x – cos 2x + sin 2x = 0


Solve the following equation:
3sin2x – 5 sin x cos x + 8 cos2 x = 2


If \[2 \sin^2 x = 3\cos x\]. where \[0 \leq x \leq 2\pi\], then find the value of x.


If \[\tan px - \tan qx = 0\], then the values of θ form a series in

 


The general solution of the equation \[7 \cos^2 x + 3 \sin^2 x = 4\] is


If \[4 \sin^2 x = 1\], then the values of x are

 


The number of values of ​x in [0, 2π] that satisfy the equation \[\sin^2 x - \cos x = \frac{1}{4}\]


General solution of \[\tan 5 x = \cot 2 x\] is


The number of values of x in the interval [0, 5 π] satisfying the equation \[3 \sin^2 x - 7 \sin x + 2 = 0\] is


Find the principal solution and general solution of the following:
sin θ = `-1/sqrt(2)`


Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°

cos 2x = 1 − 3 sin x


Solve the following equations:
2cos 2x – 7 cos x + 3 = 0


Choose the correct alternative:
If tan α and tan β are the roots of x2 + ax + b = 0 then `(sin(alpha + beta))/(sin alpha sin beta)` is equal to


Solve 2 tan2x + sec2x = 2 for 0 ≤ x ≤ 2π.


If a cosθ + b sinθ = m and a sinθ - b cosθ = n, then show that a2 + b2 = m2 + n2 


Find the general solution of the equation 5cos2θ + 7sin2θ – 6 = 0


Find the general solution of the equation sinx – 3sin2x + sin3x = cosx – 3cos2x + cos3x


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×