Question

please answer? please solution? (chapter- compound Angles) please answer?​

Answer 1

Answer:

Use calculator.

Step-by-step explanation:

Hope you are allowed to use calculator.

Answer 2

Answer:

Step-by-step explanation:

(10) We know that :

$\tan (A +B)=\frac{\tan A+\tan B }{1-\tan A\tan B}\\\\Thus\\\\\tan A + \tan B = \tan(A+B) (1-\tan A \tan B)\\Now\\\\\tan 20^\circ +\tan 40^\circ +\sqrt{3} \tan 20^\circ\tan 40^\circ\\\\= \tan 60^\circ ( 1-\tan 20^\circ\tan 40^\circ ) +\tan 60^\circ \tan 20^\circ\tan 40^\circ\\\\=\tan 60^\circ( 1-\tan 20^\circ\tan 400^\circ+\tan 20^\circ\tan 40^\circ)\\\\ \bf {=\tan 60^\circ=\sqrt{3}}$

(11)

$\cos 100^\circ\cos 40^\circ+\sin 100^\circ\sin 40^\circ\\\\=\cos (100-40)^\circ\\\\\bf {=\cos60^\circ=\frac 12}$