Question

(cos 19-sin 19)/(cos 19+sin 19)=cot 74

Answer 1

LHS = cos 19° - sin 19°cos 19° + sin 19°=cos 19°cos 19 ° - sin 19°cos 19°cos 19°cos 19° + sin 19°cos 19°=1 - tan 19°1 + tan 19°=tan 45° - tan 19°1 + tan 45° . tan ...

Answer 2

Given : $\frac{\cos 19^\circ-\sin 19^\circ }{\cos 19^\circ+\sin 19^\circ} = \cot 64^\circ$

Step-by-step explanation:

$\frac{\cos 19^\circ-\sin 19^\circ }{\cos 19^\circ+\sin 19^\circ}\\\\\text{divide by cos 19}\\\\\frac{\frac{\cos 19^\circ}{\cos 19^\circ}-\frac{\sin 19^\circ }{\cos 19^\circ}}{{\frac{\cos 19^\circ}{\cos 19^\circ}+\frac{\sin 19^\circ }{\cos 19^\circ}}}\\\\\text{we know that }\\\\\tan \theta = \frac{\sin\tehta}{\cos\theta}\\\\thus \\\\\frac{1-\tan19^\circ}{1+ \tan 19^\circ}$

$\frac{\tan 45^\circ-\tan 19^\circ }{1+\tan 45^\circ\tan 19^\circ }\\\\\tan (45^\circ - 19^\circ)\\\\tan 26^\circ\\\\\tan (90-64)\\\\\cot 64^\circ$

hence proved

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