Question

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

Answer 1

Hint: To find the value of $(1+\cot x-\csc x)(1+\tan x+\sec x)$ , we have to apply the formulas $\cot x={\displaystyle \frac{\cos x}{\sin x}},\csc x={\displaystyle \frac{1}{\sin x}},\tan x={\displaystyle \frac{\sin x}{\cos x}}$ and $\sec x={\displaystyle \frac{1}{\cos x}}$ in this expression. Then, we have to simplify and use the trigonometric and algebraic formulas including $a^2-b^2=(a+b)(a-b)$ , ${(a+b)}^2=a^2+2ab+b^2$ 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 $(1+\cot x-\csc x)(1+\tan x+\sec x)$ . We know that $\cot x={\displaystyle \frac{\cos x}{\sin x}},\csc x={\displaystyle \frac{1}{\sin x}},\tan x={\displaystyle \frac{\sin x}{\cos x}}$ and $\sec x={\displaystyle \frac{1}{\cos x}}$ . Let us substitute these results in the given trigonometric expression.
$\Rightarrow (1+{\displaystyle \frac{\cos x}{\sin x}}-{\displaystyle \frac{1}{\sin x}})(1+{\displaystyle \frac{\sin x}{\cos x}}+{\displaystyle \frac{1}{\cos x}})$
Let us take the LCM of the terms inside each bracket.
$\begin{array}{rl}& \Rightarrow ({\displaystyle \frac{1\times \sin x}{1\times \sin x}}+{\displaystyle \frac{\cos x}{\sin x}}-{\displaystyle \frac{1}{\sin x}})({\displaystyle \frac{1\times \cos x}{1\times \cos x}}+{\displaystyle \frac{\sin x}{\cos x}}+{\displaystyle \frac{1}{\cos x}})\\ & =({\displaystyle \frac{\sin x}{\sin x}}+{\displaystyle \frac{\cos x}{\sin x}}-{\displaystyle \frac{1}{\sin x}})({\displaystyle \frac{\cos x}{\cos x}}+{\displaystyle \frac{\sin x}{\cos x}}+{\displaystyle \frac{1}{\cos x}})\end{array}$
Let us add the terms inside the brackets.
$\Rightarrow \left({\displaystyle \frac{\sin x+\cos x-1}{\sin x}}\right)\left({\displaystyle \frac{\cos x+\sin x+1}{\cos x}}\right)$
We have to multiply the brackets.
$\Rightarrow {\displaystyle \frac{(\sin x+\cos x-1)(\cos x+\sin x+1)}{\sin x\cos x}}$
We can rearrange the terms inside the second bracket of the numerator as shown below.
$\Rightarrow {\displaystyle \frac{(\sin x+\cos x-1)(\sin x+\cos x+1)}{\sin x\cos x}}$
Let us group the terms as shown below.
$\Rightarrow {\displaystyle \frac{((\sin x+\cos x)-1)((\sin x+\cos x)+1)}{\sin x\cos x}}$
We can see that the numerator is of the form $a^2-b^2=(a+b)(a-b)$ . Therefore, we can write the above equation as
$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{{(\sin x+\cos x)}^2-1^2}{\sin x\cos x}}\\ & ={\displaystyle \frac{{(\sin x+\cos x)}^2-1}{\sin x\cos x}}\end{array}$
We know that ${(a+b)}^2=a^2+2ab+b^2$ . Therefore, the above equation becomes
$\Rightarrow {\displaystyle \frac{{\sin }^{2}x+2\sin x\cos x+{\cos }^{2}x-1}{\sin x\cos x}}$
We can rearrange the numerator of the above expression as
$\Rightarrow {\displaystyle \frac{{\sin }^{2}x+{\cos }^{2}x-1+2\sin x\cos x}{\sin x\cos x}}$
We know that ${\sin }^{2}x+{\cos }^{2}x=1$ . Therefore, the above expression becomes
$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{1-1+2\sin x\cos x}{\sin x\cos x}}\\ & ={\displaystyle \frac{0+2\sin x\cos x}{\sin x\cos x}}\\ & ={\displaystyle \frac{2\sin x\cos x}{\sin x\cos x}}\end{array}$
We can cancel $\sin x\cos x$ from the numerator and denominator.
$\Rightarrow {\displaystyle \frac{2\text{cancel}\sin x\cos x}{\text{cancel}\sin x\cos x}}$
We can write the result of the above simplification as
$\Rightarrow 2$
Hence, $(1+\cot x-\csc x)(1+\tan x+\sec x)=2$ .

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 ${\displaystyle \frac{1}{\cos x}}$ and $\sec x$ as ${\displaystyle \frac{1}{\sin x}}$ . 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.