Answer

Verified

418.2k+ views

**Hint:**We are given that right side is sin x tan x and left side is sec x – cos x, we will first learn how we can simplify sec x, tan x, into the simplest form then we use $\sec x=\dfrac{1}{\cos x},\tan x=\dfrac{\sin x}{\cos x}$ . We will also need the identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ . We will simplify both the right side and the left side one by one and see that they are the same .

**Complete step by step answer:**

We are asked to prove that $\sec x-\cos x$ is the same as $\sin x\tan x$ .

To prove it, we will learn how ratios are ratios connected to each other.

In our problem, we have 4 ratios so that are $\sin x,\cos x,\sec x,\tan x$ .

So, we learn how they are connected. We know that $\sin x=\dfrac{1}{\cos ecx}$ , $\cos x=\dfrac{1}{\sec x}$ or say $\sec x=\dfrac{1}{\cos x}$ , $\tan x$ is given as $\dfrac{\sin x}{\cos x}$ and lastly ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ , we can change this to get ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ or ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ as needed.

Now we will verify that $\sec x-\cos x=\sin x\tan x$ .

So, we consider the left hand side.

We have $\sec x-\cos x$ .

As we know that $\sec x=\dfrac{1}{\cos x}$

So, $\sec x-\cos x=\dfrac{1}{\cos x}-\cos x$ .By simplifying, we get –

$=\dfrac{1}{\cos x}-\dfrac{\cos x}{1}$ .

By LCM we simplify further and we get –

$=\dfrac{1-{{\cos }^{2}}x}{\cos x}$ .

We know that ${{\cos }^{2}}x+{{\sin }^{2}}x=1$

So, $1-{{\cos }^{2}}x={{\sin }^{2}}x$ .

So using the above identity, we get –

$\sec x-\cos x=\dfrac{{{\sin }^{2}}x}{\cos x}$ …………………………… (1)

So we get left side simplified as $\dfrac{{{\sin }^{2}}x}{\cos x}$

Now we consider the right side, we have $\sin x\times \tan x$ .

As $\tan x=\dfrac{\sin x}{\cos x}$ , so

$\sin x\tan x=\sin x\times \dfrac{\sin x}{\cos x}$

Simplifying, we get –

$=\dfrac{{{\sin }^{2}}x}{\cos x}$

So, $\sin x\tan x=\dfrac{{{\sin }^{2}}x}{\cos x}$ …………………………… (2)

From eq (1) and (2) we get –

$\sin x\tan x$ and $\sec x-\cos x$ are equal to one another.

**Hence $\sec x-\cos x=\sin x\tan x$**

**Note:**We can also extend the proof of the left hand side to reach to the right hand side but many time it get complicated so we normally just simplify to the simplest term possible then start working on the other side we left our left side at $\sec x-\cos x=\dfrac{{{\sin }^{2}}x}{\cos x}$ as ${{\sin }^{2}}x=\sin x\sin x$ .

So, using this, we get –

$=\dfrac{\sin x\sin x}{\cos x}$

$=\sin x\times \left( \dfrac{\sin x}{\cos x} \right)$

As $\dfrac{\sin x}{\cos x}=\tan x$

So, $=\sin x\times \tan x$

= Right Hand Side.

Hence proved.

Recently Updated Pages

Who among the following was the religious guru of class 7 social science CBSE

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Trending doubts

Write the difference between order and molecularity class 11 maths CBSE

A rainbow has circular shape because A The earth is class 11 physics CBSE

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How do you graph the function fx 4x class 9 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

What are noble gases Why are they also called inert class 11 chemistry CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Differentiate between calcination and roasting class 11 chemistry CBSE