Question

given sin x = 3/5 and cos y = -4/5 , x is an acute angle and y is and obtuse angle . find :

a)cot a + cos ec y

b)sin 2x + tan 2y

Answer 1

$$\sin ^2x +\cos^2 x=1$$

Given $\sin x=3/5, \cos x>0.$ Then

$$\cos x=\sqrt{1-\sin^2 x}=\sqrt{1-(3/5)^2}=4/5$$

Given $\cos y=-4/5, \sin y>0.$ Then

$$\sin y=\sqrt{1-\cos^2 y}=\sqrt{1-(-4/5)^2}=3/5$$

a)

$$\cot x=\dfrac{\cos x}{\sin x}=\dfrac{4/5}{3/5}=4/3$$

$$\cosec y=\dfrac{1}{\sin y}=\dfrac{1}{3/5}=5/3$$

$$\cot x+\cosec y=4/3+5/3=3$$

b)

$$\sin 2x=2\sin x\cos x=2(3/5)(4/5)=24/25$$

$$\tan y=\dfrac{\sin y}{\cos y}=\dfrac{3/5}{-4/5}=-3/4$$

$$\tan 2y=\dfrac{2\tan y}{1-\tan^2 y}=\dfrac{2(-3/4)}{1-(-3/4)^2}=-24/7$$

$$\sin 2x+\tan 2y=24/25-24/7=-432/175$$