Question

If θ is an acute angle such that $cos\Theta =\frac{3}{5}$, $\frac{sin\Thetatan\Theta-1}{2tan^{2}\Theta}=$
(a)$\frac{16}{625}$
(b)$\frac{1}{36}$
(c)$\frac{3}{160}$
(d)$\frac{160}{3}$

Answer 1

SOLUTION :  

The correct option is  (c)= 3/160

Given : cos θ = 3/5

cot θ = Base / hypotenuse = ⅗  

Base = 3 , Hypotenuse = 5  

In right angled ∆,  

Hypotenuse² = ( perpendicular)² + (Base)²

[By Pythagoras theorem]

5² = ( perpendicular)² + 3²

25 = ( perpendicular)² + 9  

(perpendicular)² = 25 - 9 = 16

perpendicular² = 16  

perpendicular = √16 = 4  

perpendicular = 4  

sin θ = perpendicular/hypotenuse = 4/5

sin θ = ⅘  

tan θ = perpendicular/base = 4/3

tan θ = 4/3

The value of : sin θ tan θ - 1  / 2tan² θ  

= (⅘ × 4/3) -1 / [ 2× (4/3)²]

= (16/15 - 1) / [2 × 16/9]

= [(16 - 15)/15] / 32/9

= (1/15) / (32/9)

= 1/15 × 9/32

= 3/ 5 × 32

sin θ tan θ - 1  / 2tan² θ  = 3/ 160

Hence, the value of  sin θ tan θ - 1  / 2tan² θ is 3/160 .

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Refer to the attachment please.