Prove that: tan 11 π 3 - 2 sin 4 π 6 - 3 4 cosec 2 π 4 + 4 cos 2 17 π 6 = 3 - 4 3 2

To prove that: tan 11 π 3 - 2 sin 4 π 6 - 3 4 cosec 2 π 4 + 4 cos 2 17 π 6 = 3 - 4 3 2 LHS = tan 11 π 3 - 2 sin 4 π 6 - 3 4 cosec 2 π 4 + 4 cos 2 17 π 6 = tan 4 π - π 3 - 2 sin π - π 3 - 3 4 × 2 + 4 cos 2 3 π - π 6 = - tan π 3 - 2 sin π 3 - 3 2 + 4 cos 2 π 6 = - 3 - 2 × 3 2 - 3 2 + 4 × 3 4 = - 2 3 - 3 2 + 3 = 3 - 4 3 2 = RHS Hence, proved.

Prove that: 3 sin π 6 sec π 3 - 4 sin 5 π 6 cot π 4 = 1

To prove that: 3 sin π 6 sec π 3 - 4 sin 5 π 6 cot π 4 = 1 L H S = 3 sin π 6 sec π 3 - 4 sin 5 π 6 cot π 4 = 3 sin π 6 sec π 3 - 4 sin π - π 6 cot π 4 We know that sine function is positive in second quadrant. = 3 × 1 2 × 2 - 4 sin π 6 × 1 = 3 - 4 × 1 2 = 1 = R H S Hence, proved.

Prove that cos ( 2 π + x ) cosec ( 2 π + x ) tan π 2 + x sec π 2 + x cos x cot ( π + x ) = 1

To prove that cos ( 2 π + x ) cosec ( 2 π + x ) tan π 2 + x sec π 2 + x cos x cot ( π + x ) = 1 L H S = cos ( 2 π + x ) cosec ( 2 π + x ) tan π 2 + x sec π 2 + x cos x cot ( π + x ) We know that all trigonometric functions are positive in first quadrant, tangent and secant functions are negative in second quadrant. So , cos ( 2 π + x ) = cos x , cosec ( 2 π + x ) = cosecx , tan π 2 + x = - cot x , sec ( π / 2 + x ) = - cosec x , cot π + x = cot x = cos x × cosec x [ - cot x ] [ - cosec x ] cos x × cot x = 1 = R H S Hence, proved.

Prove that cosec 90 ° + x + cot 450 ° + x cosec 90 ° - x + tan 180 ° - x + tan 180 ° + x + sec 180 ° - x tan 360 ° + x - sec ( - x ) = 2

To prove that cosec 90 ° + x + cot 450 ° + x cosec 90 ° - x + tan 180 ° - x + tan 180 ° + x + sec 180 ° - x tan 360 ° + x - sec ( - x ) = 2 L H S = cosec 90 ° + x + cot 450 ° + x cosec 90 ° - x + tan 180 ° - x + tan 180 ° + x + sec 180 ° - x tan 360 ° + x - sec ( - x ) = cosec 90 ° + x + cot 90 ° × 5 + x cosec 90 ° - x + tan 90 ° × 2 - x + tan 90 ° × 2 + x + sec 90 ° × 2 - x tan 90 ° × 4 + x - sec ( - x ) We know that when n is odd, cosec → sec , cot → tan and also sec ( - x ) = sec x = sec x - tan x sec x - tan x + tan x - sec x tan x - sec x = 1 + 1 = 2 = R H S

Prove that sin ( π + x ) cos π 2 + x tan 3 π 2 - x cot ( 2 π - x ) sin ( 2 π - x ) cos ( 2 π + x ) cosec ( - x ) sin 3 π 2 - x = 1

To prove that sin ( π + x ) cos π 2 + x tan 3 π 2 - x cot ( 2 π - x ) sin ( 2 π - x ) cos ( 2 π + x ) cosec ( - x ) sin 3 π 2 - x = 1 L H S = sin ( π + x ) cos π 2 + x tan 3 π 2 - x cot ( 2 π - x ) sin ( 2 π - x ) cos ( 2 π + x ) cosec ( - x ) sin 3 π 2 - x = - sin ( x ) - sin x cot x - cot ( x ) - sin ( x ) cos ( x ) - cosec ( x ) - cos x = - sin ( x ) - sin x cos ( x ) sin ( x ) - cos x sin x - sin ( x ) cos ( x ) - 1 sin ( x ) - cos x = - cos 2 x - cos 2 x = 1 = R H S Hence, proved.

Prove that 1 + cot x - sec π 2 + x 1 + cot x + sec π 2 + x = 2 cot x

To prove that 1 + cot x - sec π 2 + x 1 + cot x + sec π 2 + x = 2 cot x L H S = 1 + cot x - sec π 2 + x 1 + cot x + sec π 2 + x = { 1 + cot x - ( - cosec x ) } { 1 + cot x + ( - cosec x ) } = { ( 1 + cot x ) + ( cosec x ) } { ( 1 + cot x ) - ( cosec x ) } = ( 1 + cot x ) 2 - ( cosec x ) 2 = 1 + cot 2 x + 2 cot x - cosec 2 x (We know that 1 + cot 2 x = cosec 2 x ) = cosec 2 x + 2 cot x - cosec 2 x = 2 cot x = R H S Hence proved.

Prove that tan π 2 - x sec ( π - x ) sin ( - x ) sin ( π + x ) cot ( 2 π - x ) cosec π 2 - x = 1

To prove that tan π 2 - x sec ( π - x ) sin ( - x ) sin ( π + x ) cot ( 2 π - x ) cosec π 2 - x = 1 L H S = tan π 2 - x sec ( π - x ) sin ( - x ) sin ( π + x ) cot ( 2 π - x ) cosec π 2 - x = ( cot x ) ( - sec x ) ( - sin x ) ( - sin x ) ( - cot x ) ( sec x ) = cot x sec x sin x sin x cot x sec x = 1 = R H S Hence, proved.

E M B I B E

MATHEMATICS for CLASS XI>Chapter 5 - Trigonometric Functions>EXERCISE 5.3>Q 2

EASY

11th CBSE

IMPORTANT

Earn 100

Prove that: $\tan (- 225 °) \cot (- 405 °) - \tan (- 765 °) \cot (675 °) = 0$

Important Questions on Trigonometric Functions

EASY

11th CBSE

IMPORTANT

MATHEMATICS for CLASS XI>Chapter 5 - Trigonometric Functions>EXERCISE 5.3>Q 2

Prove that:

\cos 570 ° \sin 510 ° + \sin (- 330 °) \cos (- 390 °) = 0

EASY

11th CBSE

IMPORTANT

MATHEMATICS for CLASS XI>Chapter 5 - Trigonometric Functions>EXERCISE 5.3>Q 2

Prove that:

\tan \frac{11 π}{3} - 2 \sin \frac{4 π}{6} - \frac{3}{4} {cosec}^{2} \frac{π}{4} + 4 \cos^{2} \frac{17 π}{6} = \frac{3 - 4 \sqrt{3}}{2}

EASY

11th CBSE

IMPORTANT

MATHEMATICS for CLASS XI>Chapter 5 - Trigonometric Functions>EXERCISE 5.3>Q 2

Prove that:

3 \sin \frac{π}{6} \sec \frac{π}{3} - 4 \sin \frac{5 π}{6} \cot \frac{π}{4} = 1

EASY

11th CBSE

IMPORTANT

MATHEMATICS for CLASS XI>Chapter 5 - Trigonometric Functions>EXERCISE 5.3>Q 3

Prove that

$\frac{\cos (2 π + x) cosec (2 π + x) \tan (\frac{π}{2} + x)}{\sec (\frac{π}{2} + x) \cos x \cot (π + x)} = 1$

EASY

11th CBSE

IMPORTANT

MATHEMATICS for CLASS XI>Chapter 5 - Trigonometric Functions>EXERCISE 5.3>Q 3

Prove that $\frac{cosec (90 ° + x) + \cot (450 ° + x)}{cosec (90 ° - x) + \tan (180 ° - x)} + \frac{\tan (180 ° + x) + \sec (180 ° - x)}{\tan (360 ° + x) - \sec (- x)} = 2$

EASY

11th CBSE

IMPORTANT

MATHEMATICS for CLASS XI>Chapter 5 - Trigonometric Functions>EXERCISE 5.3>Q 3

Prove that

$\frac{\sin (π + x) \cos (\frac{π}{2} + x) \tan (\frac{3 π}{2} - x) \cot (2 π - x)}{\sin (2 π - x) \cos (2 π + x) cosec (- x) \sin (\frac{3 π}{2} - x)} = 1$