If A = 60° and B = 30°, verify that

by

If A = 60° and B = 30°, verify that

(i) sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos A cos B – sin A sin B

(iii) sin (A – B) = sin A cos B – cos A sin B

(iv) tan (A – B) = (tan A – tan B)/ (1 + tan A tan B).

Answer :

A = 60° and B = 30°

(i) sin (A + B) = sin A cos B + cos A sin B

Here

LHS = sin (A + B)

= sin (60° + 30°)

= sin 90°

= 1

RHS = sin A cos B + cos A sin B

= sin 60° cos 30° + cos 60° sin 30°

= (√3/2 × √3/2) + (½ × ½)

= ¾ + ¼

= 4/4

= 1

Hence, LHS = RHS.

(ii) cos (A + B) = cos A cos B – sin A sin B

Here

LHS = cos (A + B)

= cos (60° + 30°)

= cos 90°

= 0

RHS = cos A cos B – sin A sin B

= cos 60° cos 30° – sin 60° sin 30°

= (½× √3/2) – √3/2 + ½

= √3/4 – √3/4

= 0

Hence, LHS = RHS.

(iii) sin (A – B) = sin A cos B – cos A sin B

Here

LHS = sin (A – B)

= sin (60° – 30°)

= sin 30°

= ½

RHS = sin A cos B – cos A sin B

= sin 60° cos 30° – cos 60° sin 30°

= (√3/2 × √3/2) – (½ × ½)

= ¾ – ¼

= (3 – 1)/ 4

= 2/4

= ½

Hence, LHS = RHS.

(iv) tan (A – B) = (tan A – tan B)/ (1 + tan A tan B)

Here

LHS = tan (A – B)

= tan (60° – 30°)

= tan 30°

= 1/√3

RHS = (tan A – tan B)/ (1 + tan A tan B)

= (tan 60° – tan 30°)/(1 + tan 60° tan 30°)

Trigonometric Ratios of Standard Angles Class 9 ICSE ML Aggarwal img 9

= 2/√3 × ½

= 1/√3

Hence, LHS = RHS.

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