Question

(d) tan (45° -0)
coso - sino
coso + sine​

Answer 1

$\color{red}{\huge{\underline{\underline{GIVEN\dag}}}}$

• $\tan(45 -θ)$

$\color{red}{\huge{\underline{\underline{SOLUTION\dag}}}}$

We know that;

$\color{blue}{ {\boxed{\tan(A - B) = \frac{ \tan(A) - \tan(B) }{1 + \tan(A) \tan(B) } }}}$

$→ \tan(45-θ) = \frac{ \tan(45) - \tan(θ) }{1 + \tan(45) \tan(θ) }$

$→ \tan(45-θ) = \frac{ 1 - \tan(θ) }{1 + (1) \tan(θ) }$

$→ \tan(45-θ) = \frac{ 1 - \tan(θ) }{1 + \tan(θ) }$

$→ \tan(45-θ) = \frac{ 1 - \frac{\sinθ}{\cosθ}}{1 + \frac{\sinθ}{\cosθ} }$

$\color{orange}{{\boxed{→ \tan(45-θ) = \frac{ \cosθ - \sinθ}{\cos θ+ \sinθ }}}}$

HOPE THAT HELPS!!