Question

If $cos\Theta=\frac{3}{4}$, then find the value of 9tan²θ+ 9.

Answer 1

Answer:

The value of 9tan²θ + 9  is 16.

Step-by-step explanation:

Given : cosθ = 3/4

secθ = 1/cosθ

secθ = 1/(3/4) = 1 × (4/3) = 4/3

By using the identity , sec² θ -  tan²θ = 1

(4/3)² - tan²θ = 1

16/9 - tan²θ = 1

tan²θ = 16/9 - 1  

tan²θ = (16 - 9)/9

tan²θ = 7/9

We have to find the value of, 9tan²θ + 9 :  

9tan²θ + 9  

= 9 (7/9)+ 9

= 7 + 9

= 16

Hence, the value of 9tan²θ + 9  is 16.

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Answer 2

$\fbox{ \bold{ \large{Given :- \: \: \: }}}$

$\to \: \: \bold{cos \: \Theta=\frac{3}{4} }$

$\to$ Base = 3 units

$\to$ Hypotenuse = 4 units

$\to$ Perpendicular = $\bold{ \sqrt{ {(H) }^{2} - {(B) }^{2} } } = \bold{ \sqrt{ {(4)}^{2} - {(3)}^{2} } } = \bold{\sqrt{7} \: units }$

$\\ \bold{ \fbox{ \large{To \: Find :- \: \: \: }}}$

$\to$ 9 × tan²θ + 9

$\\ \bold{ \fbox{ \large{Solution :- \: \: \: }}}$

$\to \bold{9 \times {tan}^{2} \bold{\theta + 9}} \\ \\ \to \bold{9 \times {( \frac{P }{B }) }^{2} } \bold{ + \: 9 }\\ \\ \to \bold{ 9 \times {( \frac{ \sqrt{7} }{3} )}^{2} + 9 }\\ \\ \to \bold{9 \times \frac{7}{9} + 9} \\ \\ \to \: \: \: \: \: \: \bold{ 16 \: }$

Hence , 16 is The Required Value