In a ∆ A B C , prove the following a 2 - c 2 b 2 = sin ( A - C ) sin ( A + C ) .

Given, L . H . S = a 2 - c 2 b 2 We know that a sin A = b sin B = c sin C = k From Sine rule ⇒ a = k sin A ; b = k sin B ; c = k sin C . . . 1 Put the values of a , b , c from 1 in L . H . S , we get: = k 2 s i n 2 A - k 2 sin 2 c k 2 sin 2 B = k 2 sin 2 A - sin 2 c k 2 sin 2 B = sin 2 A - sin 2 c sin 2 B = sin ( A - C ) sin ( A + C ) sin 2 B ∵ sin 2 x - sin 2 y = sin x + y sin x - y In ∆ A B C , A + B + C = π ⇒ B = π - A + C . . . 2 = sin ( A - C ) sin ( A + C ) sin 2 ( π - ( A + C ) [From 2 ] = sin ( A - C ) sin ( A + C ) sin 2 ( ( A + C ) ∵ sin π - θ = sin θ = sin ( A - C ) sin ( A + C ) = R . H . S Hence, proved.

In a ∆ A B C , prove the following a sin ( B - C ) b 2 - c 2 = b sin ( C - A ) c 2 - a 2 = c sin ( A - B ) a 2 - b 2 .

To prove: a sin ( B - C ) b 2 - c 2 = b sin ( C - A ) c 2 - a 2 = c sin ( A - B ) a 2 - b 2 . Let us take a sin ( B - C ) b 2 - c 2 We know that: a sin A = b sin B = c sin C = k From Sine Rule . ⇒ a = k sin A ; b = k sin B ; c = k sin C . . . 1 ⇒ a sin ( B - C ) b 2 - c 2 = k sin A sin ( B - C ) k 2 sin 2 B - sin 2 C ) [From 1 ] And, A + B + C = π ⇒ A = π - B + C = K sin ( π - ( B + C ) sin ( B - C ) K 2 sin 2 B - sin 2 C = K sin B + C sin ( B - C ) K 2 sin 2 B - sin 2 C ∵ sin ( π - θ ) = sin θ = 1 k × sin 2 B - sin 2 C sin 2 B - sin 2 C ∵ sin 2 x - sin 2 y = sin x + y sin x - y = 1 k Let us take b sin ( C - A ) c 2 - a 2 ⇒ b sin ( C - A ) c 2 - a 2 = k sin B sin ( C - A ) k 2 sin 2 C - sin 2 A = sin π - ( A + C ) sin ( C - A ) k sin 2 C - sin 2 A = sin ( A + C ) sin ( C - A ) k sin 2 C - sin 2 A ∵ sin π - θ = sin θ . = sin 2 C - sin 2 A k sin 2 C - sin 2 A ∵ sin 2 x - sin 2 y = sin x + y sin x - y = 1 k Similarly , c sin ( A - B ) a 2 - b 2 = 1 k i.e a sin ( B - C ) b 2 - c 2 = b sin ( C - A ) c 2 - a 2 = c sin ( A - B ) a 2 - b 2 = 1 k Hence, proved.

In a ∆ A B C , prove the following a + b a - b = tan A + B 2 cot A - B 2 .

Given that L . H . S = a + b a - b We know that a sin A = b sin B = c sin c = 2 R From sine rule ⇒ a sin A = 2 R ⇒ a = 2 R sin A And, b sin B = 2 R ⇒ b = 2 R sin B And, c sin C = 2 R ⇒ c = 2 R sin C ∴ a + b a - b = 2 R sin A + 2 R sin B 2 R sin A - 2 R sin B = 2 R ( sin A + sin B ) 2 R ( sin A - sin B ) We know that sin x + sin y = 2 sin x + y 2 cos x - y 2 and sin x - sin y = 2 cos x + y 2 sin x - y 2 ∴ a + b a - b = 2 sin A + B 2 c os A - B 2 2 cos A + B 2 sin A - B 2 = sin A + B 2 cos A + B 2 cos A - B 2 sin A - B 2 = tan A + B 2 · cot A - B 2 = R . H . S i.e L . H . S = R . H . S Hence, proved.

E M B I B E

Mathematics Standard 11 Vol I>Chapter 3 - Trigonometry>Exercise

Tamil Nadu Board Solutions for Chapter: Trigonometry, Exercise 9: Exercise

Author:Tamil Nadu Board

Tamil Nadu Board Mathematics Solutions for Exercise - Tamil Nadu Board Solutions for Chapter: Trigonometry, Exercise 9: Exercise

Attempt the free practice questions on Chapter 3: Trigonometry, Exercise 9: Exercise with hints and solutions to strengthen your understanding. Mathematics Standard 11 Vol I solutions are prepared by Experienced Embibe Experts.

Questions from Tamil Nadu Board Solutions for Chapter: Trigonometry, Exercise 9: Exercise with Hints & Solutions

HARD

11th Tamil Nadu Board

IMPORTANT

Mathematics Standard 11 Vol I>Chapter 3 - Trigonometry>Exercise>Q 7

In a $∆ A B C,$ prove the following

$\frac{a^{2} - c^{2}}{b^{2}} = \frac{\sin (A - C)}{\sin (A + C)} .$

HARD

11th Tamil Nadu Board

IMPORTANT

Mathematics Standard 11 Vol I>Chapter 3 - Trigonometry>Exercise>Q 7

In a $∆ A B C,$ prove the following

$\frac{a \sin (B - C)}{b^{2} - c^{2}} = \frac{b \sin (C - A)}{c^{2} - a^{2}} = \frac{c \sin (A - B)}{a^{2} - b^{2}} .$

HARD

11th Tamil Nadu Board

IMPORTANT

Mathematics Standard 11 Vol I>Chapter 3 - Trigonometry>Exercise>Q 7

In a $∆ A B C,$ prove the following $\frac{a + b}{a - b} = \tan (\frac{A + B}{2}) \cot (\frac{A - B}{2}) .$

HARD

11th Tamil Nadu Board

IMPORTANT

Mathematics Standard 11 Vol I>Chapter 3 - Trigonometry>Exercise>Q 8

In a $∆ A B C,$ prove that $(a^{2} - b^{2} + c^{2}) \tan B = (a^{2} + b^{2} - c^{2}) \tan C .$

MEDIUM

11th Tamil Nadu Board

IMPORTANT

Mathematics Standard 11 Vol I>Chapter 3 - Trigonometry>Exercise>Q 9

An Engineer has to develop a triangular shaped park with a perimeter $120 m$ in a village. The park to be developed must be of maximum area. Find out the dimensions of the park. (Write the answer in $m$ ).

HARD

11th Tamil Nadu Board

IMPORTANT

Mathematics Standard 11 Vol I>Chapter 3 - Trigonometry>Exercise>Q 10

A rope of length $12$ m is given. Find the largest area of the triangle formed by this rope and find the dimensions of the triangle so formed.

HARD

11th Tamil Nadu Board

IMPORTANT

Mathematics Standard 11 Vol I>Chapter 3 - Trigonometry>Exercise>Q 11

Derive Projection formula from Law of $Sines .$

HARD

11th Tamil Nadu Board