Question

A
97. If tan A = 1 and tan B = 13; evaluate :
mi) cos A cos B - sin A sin B
(ii) sin A cos B + cos A sin B
98
A hoor​

Answer 1

$\sf{ \tan(a) = 1} \\ \implies \sf{a = \tan {}^{ - 1} (1) } \\ \implies \underline{ \sf{a = 45 \degree}}$

$\\ \sf{ \tan(b) = \sqrt{3} } \\ \implies \sf{b = \tan {}^{ - 1} ( \sqrt{3} ) } \\ \implies \underline{ \sf{b = 60 \degree}}$

• cosA . cosB - sinA . sinB

⇒cos45° × cos60° - sin45° × sin60°

$\implies \sf{ \frac{1}{ \sqrt{2} } \times \frac{1}{2} - \frac{1}{ \sqrt{2} } \times \frac{ \sqrt{3} }{2} } \\ \\ \implies \sf{ \frac{1}{2 \sqrt{2} } - \frac{ \sqrt{3} }{2 \sqrt{2} } } \\ \\ = \frac{1 - \sqrt{3} }{2 \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} } \\ \\ \implies \boxed{ \sf{ \frac{ \sqrt{2 } - \sqrt{6} }{4} }}$

• sinA . cosB + cosA . sinB

⇒sin45° × cos60° + cos45°. sin60°

$\implies \sf{ \frac{1}{ \sqrt{2} } \times \frac{1}{2} + \frac{1}{ \sqrt{2} } \times \frac{ \sqrt{3} }{2} } \\ \\ \implies \frac{1 + \sqrt{3} }{2 \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} } \\ \\ \implies \boxed{ \sf{ \frac{ \sqrt{2} + \sqrt{6} }{4} } }$