Question

Proving Identities

Verify the identity$\frac{\mathbf{1}\mathbf{+}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}}{\mathbf{1}\mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}}\mathbf{=}{\mathbf{(}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{x}\mathbf{+}\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{x}\mathbf{)}}^{\mathbf{2}}$

Short answer

The expression$\frac{\mathbf{1}\mathbf{+}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}}{\mathbf{1}\mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}}\mathbf{=}{\mathbf{(}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{x}\mathbf{+}\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{x}\mathbf{)}}^{\mathbf{2}}$ is an identity.

Step-by-step solution

Step 1. Given information

An expression$\frac{1+\sin x}{1-\sin x}={(\tan x+\sec x)}^2$

Step 2. Concept used

For such a question, split the expression into two halves, LHS and RHS. Separate them and make them easier to understand. If the outcomes are the same, it's an identity.

Step 3. Calculation

Now, simplify RHS of $\frac{1+\sin x}{1-\sin x}={(\tan x+\sec x)}^2$

localid="1651069001471" $\begin{array}{l}{(\tan x+secx)}^2={\left(\frac{\sin x}{\cos x}+\frac{1}{\cos x}\right)}^2\\ \left(\because \tan x=\frac{\sin x}{\cos x}\right)\\ =\left[\frac{{(\sin x+1)}^2}{{\cos }^{2}x}\right]\\ =\frac{{(\sin x+1)}^2}{\left(1-{\sin }^{2}x\right)}\\ \left(\because {\sin }^{2}x+{\cos }^{2}x=1\right)\\ \end{array}$

Multiply and divide by $\left(1-{\sin }^{2}x\right)$

$\begin{array}{l}=\frac{{(\sin x+1)}^2}{{\left(1-{\sin }^{2}x\right)}^2}\times \left(1-{\sin }^{2}x\right)\\ =\frac{{(\sin x+1)}^2}{{\left(1\right)}^2-{\left(\sin x\right)}^2}\\ \because a^2-b^2=(a+b)(a-b)\\ =\frac{{(\sin x+1)}^2}{(1-\sin x)(1+\sin x)}\\ =\frac{1+\sin x}{1-\sin x}\end{array}$

LHS of the equation is$\frac{1+\sin x}{1-\sin x}$

Both RHS and LHS are equal. Hence it is an identity