Đáp án:
a) Ta có: $\sin^2 a+\cos^2 a=1$
$⇒ \cos^2 a=±\left(1-\sin^2 a\right)=±\left(1-\left(\dfrac{3}{5}\right)^2\right)=±\left(1-\dfrac{9}{25}\right)=±\dfrac{16}{25}$
$⇒ \cos a=±\sqrt{\dfrac{16}{25}}=±\dfrac{4}{5}$
Mà $\dfrac{\pi}{2}<a<\pi ⇒ \cos a=-\dfrac{4}{5}$
$⇒ \tan a=\dfrac{\sin a}{\cos a}=-\dfrac{3}{4}$
$⇒ \tan\left(a-\dfrac{\pi}{4}\right)=1+\tan x=\dfrac{1}{4}$
b) Ta có: $\sin^2 a+\cos^2 a=1$
$⇒ \sin^2 a=±\left(1-\cos^2 a\right)=±\left(1-\left(\dfrac{4}{5}\right)^2\right)=±\left(1-\dfrac{16}{25}\right)=±\dfrac{9}{25}⇒\sin a=±\sqrt{\dfrac{9}{25}}=±\dfrac{3}{5}$
Mà $\dfrac{-\pi}{2}<a<0 ⇒ \sin a=-\dfrac{3}{5}$
$⇒\sin\left(a-\dfrac{\pi}{3}\right)=\sin a.\cos \dfrac{\pi}{3}-\cos a.\sin \dfrac{\pi}{3}$
$⇒ \tan\left(a+\dfrac{\pi}{6}\right)=\dfrac{\tan a+\tan \dfrac{\pi}{6}}{1+\tan a.\tan \dfrac{\pi}{6}}$
