Question

How do you verify ${\text{tan(}}\theta {\text{)cosec(}}\theta {\text{) = sec(}}\theta {\text{)}}$ ?

Answer 1

Hint: In this question, we are asked to verify the given statement. First we have to expand the terms using the properties of trigonometric functions and its expressions. And then we need to equate or interpret the expression of the LHS in such a way that we can get the RHS. By this we get the RHS.

Formula used: The properties of trigonometric functions used:
${\text{tan(}}\theta {\text{) = }}\dfrac{{\sin (\theta )}}{{\cos (\theta )}}$
${\text{cosec(}}\theta {\text{) = }}\dfrac{1}{{\sin (\theta )}}$
${\text{sec(}}\theta {\text{) = }}\dfrac{1}{{\cos (\theta )}}$

Complete step-by-step solution:
We are asked to verify that the given LHS is equal is LHS i.e. ${\text{tan(}}\theta {\text{)cosec(}}\theta {\text{) = sec(}}\theta {\text{)}}$
For this, we will use the properties of trigonometric functions.
From the properties of trigonometric functions, we have
$ \Rightarrow {\text{tan(}}\theta {\text{) = }}\dfrac{{\sin (\theta )}}{{\cos (\theta )}}$
$ \Rightarrow {\text{cosec(}}\theta {\text{) = }}\dfrac{1}{{\sin (\theta )}}$
Now replacing ${\text{tan(}}\theta {\text{) }}$ by ${\text{ }}\dfrac{{\sin (\theta )}}{{\cos (\theta )}}$ and ${\text{cosec(}}\theta {\text{) }}$ by $\dfrac{1}{{\sin (\theta )}}$, we get
LHS =${\text{tan(}}\theta {\text{)cosec(}}\theta {\text{)}} = \dfrac{{\sin (\theta )}}{{\cos (\theta )}} \times \dfrac{1}{{\sin (\theta )}}{\text{ }}$
As there is a$\sin (\theta )$ in both the numerator and denominator, we can cancel it to simplify. And it will become like this,
$ \Rightarrow {\text{tan(}}\theta {\text{)cosec(}}\theta {\text{)}} = \dfrac{1}{{\cos (\theta )}}$
Now we use the formula ${\text{sec(}}\theta {\text{) = }}\dfrac{1}{{\cos (\theta )}}$, we have to replace ${\text{ }}\dfrac{1}{{\cos (\theta )}}$by ${\text{sec(}}\theta {\text{) }}$.
$ \Rightarrow {\text{tan(}}\theta {\text{)cosec(}}\theta {\text{)}} = \sec (\theta )$
And we get the RHS,
Hence proved.

Note: In this question we have alternative method as follows:
Alternative method:
In this question, we need to verify that${\text{tan(}}\theta {\text{)cosec(}}\theta {\text{)}} = \sec (\theta )$.
We can verify this by solving RHS too.
We all know the properties of trigonometric functions, from that we can say
$ \Rightarrow \sec (\theta ) = \dfrac{1}{{\cos (\theta )}}$
Therefore, by replacing $\sec (\theta )$ by $\dfrac{1}{{\cos (\theta )}}$, we get
$ \Rightarrow \sec (\theta ) = \dfrac{1}{{\cos (\theta )}}$
Now, we have to multiply and divide it by $\sin (\theta )$,
$ \Rightarrow \sec (\theta ) = \dfrac{1}{{\cos (\theta )}} \times \dfrac{{\sin (\theta )}}{{\sin (\theta )}}$
Making it as an expression we get,
$ \Rightarrow \sec (\theta ) = \dfrac{{\sin (\theta )}}{{\cos (\theta )\sin (\theta )}}$
Now we need to split this as$\dfrac{{\sin (\theta )}}{{\cos (\theta )}} \times \dfrac{1}{{\sin (\theta )}}$,
$ \Rightarrow \sec (\theta ) = \dfrac{{\sin (\theta )}}{{\cos (\theta )}} \times \dfrac{1}{{\sin (\theta )}}$
As we know that,
$ \Rightarrow {\text{tan(}}\theta {\text{) = }}\dfrac{{\sin (\theta )}}{{\cos (\theta )}}$
$ \Rightarrow {\text{cosec(}}\theta {\text{) = }}\dfrac{1}{{\sin (\theta )}}$
We will replace $\dfrac{{\sin (\theta )}}{{\cos (\theta )}}$ by ${\text{tan(}}\theta {\text{)}}$ and $\dfrac{1}{{\sin (\theta )}}$ by ${\text{cosec(}}\theta {\text{)}}$ to get the answer.
Therefore it becomes, LHS =$\sec (\theta ) = \tan (\theta )\cos {\text{ec}}(\theta )$.
Hence the given statement is verified.