Question

√1+sinx÷1-sinx


pl help me ​

Answer 1

Answer:

secx+tanx

Step-by-step explanation:

√1+sinx/1-sinx

=√(1+sinx)(1+sinx)/(1-sinx)(1+sinx)

=√(1+sinx)²/1-sin²x

=√(1+sinx)²/cos²x     (∵x ix acute both sinx and cosx are positive )

=1+sinx/cosx

=secx+tanx

∴Option (a) is correct.

secx+tanx

Answer 2

Trigonometric functions

We are asked to find the exact value of the following equation, if $x$ is acute:

$\longrightarrow \sqrt{\dfrac{1 + \sin(x)}{1 - \sin(x)}}$

We will be using the following identities to solve the problem.

$\boxed{\begin{array}{l}\qquad\;\quad\bf{\dag}\;\textsf{\textbf{\underline{\;IDENTITIES\;}}}\\\\\bull\;\;1-\sin(x) \cdot 1+\sin(x)=\cos^2(x)\\\\\bull\;\;\dfrac{1}{\cos(x)}=\sec(x)\\\\\bull\;\;\dfrac{\sin(x)}{\cos(x)}=\tan(x)\end{array}}$

Now, consider the equation;

$\implies \sqrt{\dfrac{1 + \sin(x)}{1 - \sin(x)}}$

$\implies \sqrt{\dfrac{1 + \sin(x)}{1 - \sin(x)} \cdot \dfrac{1 + \sin(x)}{1 + \sin(x)}}$

$\implies \sqrt{\dfrac{\left(1 + \sin(x)\right)\left(1 + \sin(x)\right)}{\left(1 - \sin(x)\right)\left(1 + \sin(x)\right)}}$

$\implies \sqrt{\dfrac{\left(1 + \sin(x)\right)^{2}}{\cos^{2}(x)}}$

$\implies \dfrac{1 + \sin(x)}{\cos(x)}$

$\implies \dfrac{1}{\cos(x)} + \dfrac{\sin(x)}{\cos(x)}$

$\implies \boxed{\sec(x) + \tan(x)}$

Hence, option (a) is correct.