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# What is $\left( 1+\cot x-\csc x \right)\left( 1+\tan x+\sec x \right)$ equal to?(a) 1(b) 2(c) $\sin x$(d) $\cos x$

Last updated date: 19th Jul 2024
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Hint: To find the value of $\left( 1+\cot x-\csc x \right)\left( 1+\tan x+\sec x \right)$ , we have to apply the formulas $\cot x=\dfrac{\cos x}{\sin x},\csc x=\dfrac{1}{\sin x},\tan x=\dfrac{\sin x}{\cos x}$ and $\sec x=\dfrac{1}{\cos x}$ in this expression. Then, we have to simplify and use the trigonometric and algebraic formulas including ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ , ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ and ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ . Then, we have to simplify the expression.

Complete step by step solution:
We have to find the value of $\left( 1+\cot x-\csc x \right)\left( 1+\tan x+\sec x \right)$ . We know that $\cot x=\dfrac{\cos x}{\sin x},\csc x=\dfrac{1}{\sin x},\tan x=\dfrac{\sin x}{\cos x}$ and $\sec x=\dfrac{1}{\cos x}$ . Let us substitute these results in the given trigonometric expression.
$\Rightarrow \left( 1+\dfrac{\cos x}{\sin x}-\dfrac{1}{\sin x} \right)\left( 1+\dfrac{\sin x}{\cos x}+\dfrac{1}{\cos x} \right)$
Let us take the LCM of the terms inside each bracket.
\begin{align} & \Rightarrow \left( \dfrac{1\times \sin x}{1\times \sin x}+\dfrac{\cos x}{\sin x}-\dfrac{1}{\sin x} \right)\left( \dfrac{1\times \cos x}{1\times \cos x}+\dfrac{\sin x}{\cos x}+\dfrac{1}{\cos x} \right) \\ & =\left( \dfrac{\sin x}{\sin x}+\dfrac{\cos x}{\sin x}-\dfrac{1}{\sin x} \right)\left( \dfrac{\cos x}{\cos x}+\dfrac{\sin x}{\cos x}+\dfrac{1}{\cos x} \right) \\ \end{align}
Let us add the terms inside the brackets.
$\Rightarrow \left( \dfrac{\sin x+\cos x-1}{\sin x} \right)\left( \dfrac{\cos x+\sin x+1}{\cos x} \right)$
We have to multiply the brackets.
$\Rightarrow \dfrac{\left( \sin x+\cos x-1 \right)\left( \cos x+\sin x+1 \right)}{\sin x\cos x}$
We can rearrange the terms inside the second bracket of the numerator as shown below.
$\Rightarrow \dfrac{\left( \sin x+\cos x-1 \right)\left( \sin x+\cos x+1 \right)}{\sin x\cos x}$
Let us group the terms as shown below.
$\Rightarrow \dfrac{\left( \left( \sin x+\cos x \right)-1 \right)\left( \left( \sin x+\cos x \right)+1 \right)}{\sin x\cos x}$
We can see that the numerator is of the form ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . Therefore, we can write the above equation as
\begin{align} & \Rightarrow \dfrac{{{\left( \sin x+\cos x \right)}^{2}}-{{1}^{2}}}{\sin x\cos x} \\ & =\dfrac{{{\left( \sin x+\cos x \right)}^{2}}-1}{\sin x\cos x} \\ \end{align}
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ . Therefore, the above equation becomes
$\Rightarrow \dfrac{{{\sin }^{2}}x+2\sin x\cos x+{{\cos }^{2}}x-1}{\sin x\cos x}$
We can rearrange the numerator of the above expression as
$\Rightarrow \dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x-1+2\sin x\cos x}{\sin x\cos x}$
We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ . Therefore, the above expression becomes
\begin{align} & \Rightarrow \dfrac{1-1+2\sin x\cos x}{\sin x\cos x} \\ & =\dfrac{0+2\sin x\cos x}{\sin x\cos x} \\ & =\dfrac{2\sin x\cos x}{\sin x\cos x} \\ \end{align}
We can cancel $\sin x\cos x$ from the numerator and denominator.
$\Rightarrow \dfrac{2\require{cancel}\cancel{\sin x\cos x}}{\require{cancel}\cancel{\sin x\cos x}}$
We can write the result of the above simplification as
$\Rightarrow 2$
Hence, $\left( 1+\cot x-\csc x \right)\left( 1+\tan x+\sec x \right)=2$ .

So, the correct answer is “Option b”.

Note: Students must be thorough with the formulas of trigonometric functions. They have a chance of making a mistake by writing the formula for $\csc x$ as $\dfrac{1}{\cos x}$ and $\sec x$ as $\dfrac{1}{\sin x}$ . Also, students may be get confused with the formula ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ by writing the value of ${{\sin }^{2}}x+{{\cos }^{2}}x$ as -1. They must also be thorough with algebraic identities.