Question

If tan A = 1 and tan B =$\sqrt{3}$
; evaluate :
(i) cos A cos B - sin A sin B
(ii) sin A cos B + cos A sin B​

Answer 1

Question :-

If tan A = 1 and tan B =$\sqrt{3}$

; evaluate :

• (i) cos A cos B - sin A sin B

• (ii) sin A cos B + cos A sin B

Given :-

• tanA = 1

• tanB = $\sqrt{3}$

To find :-

• (i) cos A cos B - sin A sin B

• (ii) sin A cos B + cos A sin B

Solution :-

$\large \implies \tan(A) = 1 \bf \: \: \: \: (given)$

$\\ \implies \large \cancel{tan}(A) = \cancel{tan}45^{\circ}$

$\\ \large \implies \: A = 45^{\circ}$

$\\ \large \implies \tan(B) = \sqrt{3} \bf \: \: \: (given)$

$\large \implies \ \cancel{tan}(B) = \cancel{tan}60^{\circ}$

$\\ \large \implies \: B = 60^{\circ}$

$\\ (1) \: \large \cos A \times \cos B - \sin A × \sin B$

$\\ \large \implies\cos45°× \cos 60°- \sin 45°- \sin 60°$

$\large = \dfrac{1}{ \sqrt{2} } \times \dfrac{1}{2} - \dfrac{1}{ \sqrt{2} } \times \dfrac{ \sqrt{3} }{2}$

$\\ \large = \frac{1}{2 \sqrt{2} } - \frac{ \sqrt{3} }{2 \sqrt{2} }$

$\\ \large = \frac{1 - \sqrt{3} }{2 \sqrt{2} }$

Now rationalise the denominator

$\\ \large = \frac{1 - \sqrt{3} }{2 \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} }$

$\\ = \large \frac{ \sqrt{2} - \sqrt{6} }{4}$

$\\ (2) \: \large \ \sin A \times \cos B + \ \cos A × \sin B$

$\\ \large \implies\ \sin 45°× \cos 60°- \ \cos 45°- \sin 60°$

$\large = \dfrac{1}{ \sqrt{2} } \times \dfrac{1}{2} + \dfrac{1}{ \sqrt{2} } \times \dfrac{ \sqrt{3} }{2}$

$\\ \large = \frac{1}{2 \sqrt{2} } + \frac{ \sqrt{3} }{2 \sqrt{2} }$

$\\ \large = \frac{1 + \sqrt{3} }{2 \sqrt{2} }$

Now rationalise the denominator

$\\ \large = \frac{1 + \sqrt{3} }{2 \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} }$

$\\ = \large \frac{ \sqrt{2} + \sqrt{6} }{4}$