Question

if 3sec^2A+8=10sec^2A,find the values of tanA.

Answer 1

Answer:

tana= 1/√7

Step-by-step explanation:

I hope this will help you

Answer 2

Answer:

$\frac{1}{\sqrt{7} }$

Step-by-step explanation:

3sec²A + 8 = 10sec²A

⇒ 10sec²A - 3sec²A = 8

⇒ sec²A(10-3) = 8

⇒ 7sec²A = 8

⇒ sec²A = 8/7

⇒ secA = $\sqrt{ \frac{8}{7} }$ = $\frac{\sqrt{8} }{\sqrt{7} }$

Wkt, secФ = 1/cosФ

Also, tanФ = sinФ/cosФ

∴ tanФ = secФ x sinФ

Now,

$\frac{1}{cosA} = \sqrt{ \frac{8}{7} }$

⇒ $cosA = \sqrt{ \frac{7}{8} }$

Wkt cosФ = $\frac{adjacent}{hypotenuse}$

∴ Adjacent = √7

Hypotenuse = √8

By using Pythogoras theorem,

(Hypotenuse)² = (Opposite)² + (Adjacent)²

⇒ (√8)² = (Opposite)² + (√7)²

⇒ (Opposite)² = (√8)² - (√7)²

⇒ (Opposite)² = 8 - 7

⇒ (Opposite)² = 1

∴ Opposite = √1 = 1

Wkt sinФ = $\frac{opposite}{hypotenuse}$

∴ sinA = $\frac{1}{\sqrt{8} }$

Now, tanA = sinA x secA

⇒ tanA = $\frac{1}{\sqrt{8} }$ x $\frac{\sqrt{8} }{\sqrt{7} }$

∴ tanA = $\frac{1}{\sqrt{7} }$

Hope this helped!