Question

DISCUSS: Equations That Are Not Identities How can you tell if an equation is not an identity? Show that the following equations are not identities.

$\begin{array}{l}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{2}\mathit{x}\mathbf{=}\mathbf{2}\mathbf{}\mathbf{sin}\mathbf{}\mathit{x}\\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}\mathbf{sin}\mathbf{(}\mathit{x}\mathbf{+}\mathit{y}\mathbf{)}\mathbf{=}\mathbf{sin}\mathbf{}\mathit{x}\mathbf{+}\mathbf{sin}\mathbf{}\mathit{y}\\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}{\mathbf{sec}}^{\mathbf{2}}\mathbf{}\mathit{x}\mathbf{+}{\mathbf{csc}}^{\mathbf{2}}\mathbf{}\mathit{x}\mathbf{=}\mathbf{1}\\ \mathbf{\left(}\mathbf{d}\mathbf{\right)}\mathbf{}\frac{\mathbf{1}}{\mathbf{sin}\mathbf{}\mathit{x}\mathbf{+}\mathbf{cos}\mathbf{}\mathit{x}}\mathbf{=}\mathbf{csc}\mathbf{}\mathit{x}\mathbf{+}\mathbf{sec}\mathbf{}\mathit{x}\end{array}$

Short answer

(a)$\sin 2x\doteq 2\sin x$ is not an identity.

(b)$\sin (x+y)=\sin x+\sin y$ is not an identity.

(c)${\sec }^{2}x+{\csc }^{2}x=1$ is not an identity.

(d)$\frac{1}{\sin x+\cos x}=\csc x+\sec x$ is not an identity.

Step-by-step solution

Part a. Step 1. Given information.

The given equation,$\sin 2x=2\sin x$

Part a. Step 2. Take random values of x and show that it in not an identity.

For,$\sin 2x=2\sin x$

Take,$x=\frac{\pi }{2}$

Put the value of $x$in the equation:-

LHS:

$\begin{array}{l}\sin 2x=\sin 2*\frac{\pi }{2}\\ \sin 2x=\sin \pi \\ \sin 2x=0\end{array}$

RHS:

$\begin{array}{l}2\sin x=2\sin \frac{\pi }{2}\\ 2\sin x=2\\ \text{LHS}\ne \text{RHS}\end{array}$

Let take$x=\frac{\pi }{4}$

LHS:

$\begin{array}{l}\sin 2x=\sin 2*\frac{\pi }{4}\\ \sin 2x=\sin \frac{\pi }{2}\\ \sin 2x=1\end{array}$

RHS:

$\begin{array}{r}2\sin x=2\sin \frac{\pi }{4}\\ 2\sin x=2*\frac{1}{\sqrt{2}}\end{array}$

$\text{LHS}\ne \text{RHS}$

Part b. Step 1. Given information.

The given equation,$\sin (x+y)=\sin x+\sin y$

Part b. Step 2. we have to put the random values in LHS and RHS. If they are not equal then we can say that given equation is not an identity.

Identity is a equation that is true for all values
In this these question we have to simply show that two sides are not equal for a particular values.

Given equation is, $\sin (x+y)=\sin x+\sin y$

Let take, $x=\frac{\pi }{6}$and $y=\frac{\pi }{6}$

Simply, Put in the LHS $=\sin (x+y):-$

$\begin{array}{l}\sin (x+y)=\sin \left(\frac{\pi }{6}+\frac{\pi }{6}\right)\\ \sin (x+y)=\sin \left(\frac{\pi }{3}\right)\\ \sin (x+y)=\frac{\sqrt{3}}{2}\end{array}$

Now, Put in the$\mathrm{RHS}=\sin x+\sin y:-$

$\begin{array}{l}\sin x+\sin y=\sin \left(\frac{\pi }{6}\right)+\sin \left(\frac{\pi }{6}\right)\\ \sin x+\sin y=\frac{1}{2}+\frac{1}{2}\\ \sin x+\sin y=1\end{array}$

As, we can see that at a particular values of $x$ and $y$. LHS and RHS is not equal.

It means that$L.H.S.\ne R.H.S$

Therefore, following given equation is not an identity.

Part c. Step 1. Given information.

The given equation,${\sec }^{2}x+{\csc }^{2}x=1$

Part c. Step 2. Simply, we have to put the random values in LHS and RHS.

If they are not equal then we can say that given equation is not an identity.

Identity is a equation that is true for all values
In this these question we have to simply show that two sides are not equal for a particular values.

Given equation is, ${\sec }^{2}x+{\csc }^{2}x=1$

Let take, $x=\frac{\pi }{4}$

Simply, Put in the LHS $={\sec }^{2}x+{\csc }^{2}x$.

width="278" height="97" role="math">sec2x+csc2x=sec2π4+csc2π4sec2x+csc2x=2+2sec2x+csc2x=4

As, we can see that at a particular value of $x$, LHSand RHS is not equal.

It means that $LHS\ne RHS$.
Therefore, following given equation is not an identity.

Part d. Step 1. Given information.

The given equation,$\frac{1}{\sin x+\cos x}=\csc x+\sec x$

Part d. Step 2. Simply, we have to put the random values in LHS and RHS.

If they are not equal then we can say that given equation is not an identity.

Identity is a equation that is true for all values
In this these question we have to simply show that two sides are not equal for a particular values.

Given equation is, $\frac{1}{\sin x+\cos x}=\csc x+\sec x$.

Let take, $x=\frac{\pi }{4}$

Simply, Put in the LHS $=\frac{1}{\sin x+\cos x}:-$

$\begin{array}{l}\frac{1}{\sin x+\cos x}=\frac{1}{\sin \left(\frac{\pi }{4}\right)+\cos \left(\frac{\pi }{4}\right)}\\ \frac{1}{\sin x+\cos x}=\frac{1}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}\\ \frac{1}{\sin x+\cos x}=\frac{1}{\frac{2}{\sqrt{2}}}\\ \frac{1}{\sin x+\cos x}=\frac{\sqrt{2}}{2}\end{array}$

$\frac{1}{\sin x+\cos x}=\frac{1}{\sqrt{2}}$

Now, Put in the RHS $\csc x+\sec x:-$

$\begin{array}{l}\csc x+\sec x=\csc \left(\frac{\pi }{4}\right)+\sec \left(\frac{\pi }{4}\right)\\ \csc x+\sec x=\sqrt{2}+\sqrt{2}\\ \csc x+\sec x=2\sqrt{2}\end{array}$

As, we can see that at a particular values of x. LHS and RHS is not equal. It means that$L.H.S.\ne R.H.S$

Therefore, following given equation is not an identity.