Question

sec theta- tan theta = 1/root 3 then the value of sec theta+ tan theta??

Answer 1

Hey !!!

Sec¢ - tan¢ = 1/√3 (given)

we know that sec²¢ - tan²¢ = 1

sec²¢ - tan²¢ =( sec¢ - tan¢ )(sec¢ + tan¢ ) = 1
like a²-b² = (a+b)(a-b)

(1/√3)(sec¢ + tan¢) = 1 [ •°•sec¢ - tan¢ = 1/√3 given )

so .. sec¢ + tan¢ = √3 Answer

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Hope it helps you !!!


@Rajukumar111

Answer 2

The value of $\sec \theta+\tan \theta$ is $\sqrt{3}$.

Step-by-step explanation:

We have,

$\sec \theta-\tan \theta=\dfrac{1}{\sqrt{3}}$       .....(1)

To find, the value of $\sec \theta+\tan \theta=?$

We know that,

$\sec^{2} \theta-\tan^{2} \theta=1$

∴ $(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)=1$

[ ∵ $a^{2} -b^{2} =(a-b)(a+b)$]

Using (1), we get

$(\dfrac{1}{\sqrt{3}} )(\sec \theta+\tan \theta)=1$

⇒ $\sec \theta+\tan \theta=\sqrt{3}$

Hence, the value of $\sec \theta+\tan \theta$ is $\sqrt{3}$.