मराठी

Prove That: Cos ( 2 π + X ) C O S E C ( 2 π + X ) Tan ( π / 2 + X ) Sec ( π / 2 + X ) Cos X Cot ( π + X )

Advertisements
Advertisements

प्रश्न

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

 

Advertisements

उत्तर

 LHS = \[\frac{\cos\left( 2\pi + x \right) cosec\left( 2\pi + x \right) \tan\left( \frac{\pi}{2} + x \right)}{\sec \left( \frac{\pi}{2} + x \right) \cos x \cot \left( \pi + x \right)}\]
\[ = \frac{\cos x cosec x \left[ - \cot x \right]}{\left[ - cosec x \right]\cos x \cot x} \]
\[ = \frac{- \cos x cosec x \cot x}{- cosec x cos x \cot x}\]
\[ = 1\]
= RHS
Hence proved.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Trigonometric Functions - Exercise 5.3 [पृष्ठ ३९]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
पाठ 5 Trigonometric Functions
Exercise 5.3 | Q 3.1 | पृष्ठ ३९

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Find the principal and general solutions of the equation `tan x = sqrt3`


If \[\tan x = \frac{b}{a}\] , then find the values of \[\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}\].


Prove the:
\[ \sqrt{\frac{1 - \sin x}{1 + \sin x}} + \sqrt{\frac{1 + \sin x}{1 - \sin x}} = - \frac{2}{\cos x},\text{ where }\frac{\pi}{2} < x < \pi\]


If \[T_n = \sin^n x + \cos^n x\], prove that \[\frac{T_3 - T_5}{T_1} = \frac{T_5 - T_7}{T_3}\]

 


Prove that:

\[\sin\frac{8\pi}{3}\cos\frac{23\pi}{6} + \cos\frac{13\pi}{3}\sin\frac{35\pi}{6} = \frac{1}{2}\]

 


Prove that:
\[\sin\frac{13\pi}{3}\sin\frac{8\pi}{3} + \cos\frac{2\pi}{3}\sin\frac{5\pi}{6} = \frac{1}{2}\]


Prove that:

\[\sin\frac{10\pi}{3}\cos\frac{13\pi}{6} + \cos\frac{8\pi}{3}\sin\frac{5\pi}{6} = - 1\]

If sec \[x = x + \frac{1}{4x}\], then sec x + tan x = 

 

If x = r sin θ cos ϕ, y = r sin θ sin ϕ and r cos θ, then x2 + y2 + z2 is independent of


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

 


If sec x + tan x = k, cos x =


Find the general solution of the following equation:

\[\cos x = - \frac{\sqrt{3}}{2}\]

Find the general solution of the following equation:

\[\tan x = - \frac{1}{\sqrt{3}}\]

Find the general solution of the following equation:

\[\sqrt{3} \sec x = 2\]

Find the general solution of the following equation:

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

Solve the following equation:
\[\sin^2 x - \cos x = \frac{1}{4}\]


Solve the following equation:

\[2 \sin^2 x + \sqrt{3} \cos x + 1 = 0\]

Solve the following equation:

\[\cos 4 x = \cos 2 x\]

Solve the following equation:

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

Solve the following equation:

\[\sin x + \sin 2x + \sin 3x + \sin 4x = 0\]

Solve the following equation:

\[\sqrt{3} \cos x + \sin x = 1\]


Solve the following equation:
\[\sec x\cos5x + 1 = 0, 0 < x < \frac{\pi}{2}\]


Solve the following equation:
 cosx + sin x = cos 2x + sin 2x

 


The smallest value of x satisfying the equation

\[\sqrt{3} \left( \cot x + \tan x \right) = 4\] is 

If \[\cos x + \sqrt{3} \sin x = 2,\text{ then }x =\]

 


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


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


The equation \[3 \cos x + 4 \sin x = 6\] has .... solution.


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


Find the principal solution and general solution of the following:
tan θ = `- 1/sqrt(3)`


Choose the correct alternative:
If tan 40° = λ, then `(tan 140^circ - tan 130^circ)/(1 + tan 140^circ *  tan 130^circ)` =


Choose the correct alternative:
If cos pθ + cos qθ = 0 and if p ≠ q, then θ is equal to (n is any integer)


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


Choose the correct alternative:
If f(θ) = |sin θ| + |cos θ| , θ ∈ R, then f(θ) is in the interval


In a triangle ABC with ∠C = 90° the equation whose roots are tan A and tan B is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×