Question

if sinθ=3/5,then prove the value of sec^2θ+tan^2θ=34/16​

Answer 1

Solution:

Given that,

→ Sin Ф = 3/5

From the diagram, we get,

→ Cos Ф = 4/5

→ Tan Ф = 3/4

Therefore, Sec Ф = (1 / Cos Ф) which is given as:

→ Sec Ф = 5/4

Therefore Sec²Ф + Tan²Ф is given as:

$\rightarrow (\dfrac{5}{4})^2 + (\dfrac{3}{4})^2 = Sec^2\theta + Tan^2\theta\\\\\\\text{Consider LHS we get,}\\\\\\\rightarrow \dfrac{25}{16} + \dfrac{9}{16} \implies \dfrac{34}{16}\\\\\\\text{Hence, we get,}\\\\\boxed{\rightarrow Sec^2\theta + Tan^2\theta = \dfrac{34}{16}}\\\\\\\textbf{Hence Proved}$

Answer 2

$\huge \mathtt{ \fbox{Solution :)}}$

Given ,

Sinθ = 3/5 = P/H

By phythagores theorem ,

(5)² = (B)² + (3)²

25 = (B)² + 9

(B)² = 25 - 9

(B)² = 16

B = 4 units ,

[ ignore the negative value of B because length can't be negative ]

Hence , base , perpendicular and hypotenuse of the right angled triangle are 4 units , 3 units and 5 units

Now ,

➡(Sec)²θ + (Tan)²θ

➡(H/B)² + (P/B)²

➡(5/4)² + (3/4)²

➡(25 + 9)/16

➡34/16

Hence , (Sec)²θ + (Tan)²θ = 34/16

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