Question

If $\sin x+\cos x+\tan x+\cot x+\sec x+\operatorname{cosec}x=7$ and $\sin 2x=a-b\sqrt{c}$, then $a-b+2c$ is
a. 0
b. 14
c. 28
d. 42

Answer 1

This is a question related to the trigonometric ratios. So, here, we will convert all the trigonometric functions in terms of sin x and cos x and after that convert them to sin 2x. We will solve using the following relations,
$\begin{aligned} & \tan x=\dfrac{\sin x}{\cos x},\cot x=\dfrac{\cos x}{\sin x},\sec x=\dfrac{1}{\cos x},\operatorname{cosec}x=\dfrac{1}{\sin x} \\\ & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\\ & {{\left( \sin x+\cos x \right)}^{2}}=1+\sin 2x \\\ & \sin 2x=2\sin x\cos x \\\ \end{aligned}$

Complete step by step answer:
Now, we have been given in the question that,
$\sin x+\cos x+\tan x+\cot x+\sec x+\operatorname{cosec}x=7$
We will express the above equation in terms of sin x and cos x, so for that we can write $\tan x=\dfrac{\sin x}{\cos x},\cot x=\dfrac{\cos x}{\sin x},\sec x=\dfrac{1}{\cos x},\operatorname{cosec}x=\dfrac{1}{\sin x}$
So, on substituting these values in the equation, we will get the equation as,
$\left( \sin x+\cos x \right)+\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}+\dfrac{1}{\cos x}+\dfrac{1}{\sin x}=7$
Now, on taking the LCM of the terms, $\dfrac{\sin x}{\cos x}\operatorname{and}\dfrac{\cos x}{\sin x}$ and also $\dfrac{1}{\cos x}\operatorname{and}\dfrac{1}{\sin x}$, we will get,
$\left( \sin x+\cos x \right)+\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\sin x\cos x}+\dfrac{\sin x+\cos x}{\sin x\cos x}=7$
On taking (sin x + cos x) common from the terms $\left( \sin x+\cos x \right)\operatorname{and}\dfrac{\sin x+\cos x}{\sin x\cos x}$, we get,
$\left( \sin x+\cos x \right)\left[ 1+\dfrac{1}{\sin x\cos x} \right]+\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\sin x\cos x}=7$
Now, we know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, so we can write,
$\left( \sin x+\cos x \right)\left[ 1+\dfrac{1}{\sin x\cos x} \right]+\dfrac{1}{\sin x\cos x}=7$
Now, we know that ${{\left( \sin x+\cos x \right)}^{2}}=1+\sin 2x$, so we can write $\left( \sin x+\cos x \right)=\sqrt{1+\sin 2x}$.
Also, we know that $\sin 2x=2\sin x\cos x$, so we can write $\sin x\cos x=\dfrac{\sin 2x}{2}$.
So, on substituting these values in our equation, we get,
$\left( \sqrt{1+\sin 2x} \right)\left[ 1+\dfrac{2}{\sin 2x} \right]+\dfrac{2}{\sin 2x}=7$
Now, let us consider sin 2x = y. So, we get,
$\left( \sqrt{1+y} \right)\left[ 1+\dfrac{2}{y} \right]+\dfrac{2}{y}=7$
On transposing $\dfrac{2}{y}$ from LHS to RHS, we get,
$\sqrt{1+y}\left[ 1+\dfrac{2}{y} \right]=7-\dfrac{2}{y}$
On taking the LCM of the terms in the RHS and the terms inside the bracket in the LHS, we can further write it as,
$\sqrt{1+y}\left[ \dfrac{y+2}{y} \right]=\dfrac{7y-2}{y}$
Now, on taking $\left[ \dfrac{y+2}{y} \right]$ from LHS to RHS, we get,

& \sqrt{1+y}=\dfrac{7y-2}{y}\times \dfrac{y}{y+2} \\\ & \sqrt{1+y}=\dfrac{7y-2}{y+2} \\\ \end{aligned}$$ On squaring both the sides, we get, $y+1=\dfrac{{{\left( 7y-2 \right)}^{2}}}{{{\left( y+2 \right)}^{2}}}$ On cross multiplying, we get, $\left( y+1 \right){{\left( y+2 \right)}^{2}}={{\left( 7y-2 \right)}^{2}}$ Now, we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ so we can write, $\left( y+1 \right)\left( {{y}^{2}}+4+4y \right)=\left( 49{{y}^{2}}+4-28y \right)$ On opening the brackets, we get, ${{y}^{3}}+4{{y}^{2}}+4y+{{y}^{2}}+4y+4=49{{y}^{2}}+4-28y$ On solving further, we get, ${{y}^{3}}-44{{y}^{2}}+36y=0$ On taking y common from all the terms, we get, $y\left( {{y}^{2}}-44y+36 \right)=0$ Now, we will substitute back the value of y as sin 2x, so we get, $\sin 2x\left( {{\sin }^{2}}2x-44\sin 2x+36 \right)=0$ So, we can write, sin 2x = 0 But this is not possible, as cot x and cosec x are not defined. Next, we can write, ${{\sin }^{2}}2x-44\sin 2x+36=0$ We will find the value of sin 2x, by using the quadratic equation, which is $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, and here we have a = 1, b = -44 and c = 36. So, we will get, $\begin{aligned} & \sin 2x=\dfrac{-\left( -44 \right)\pm \sqrt{{{\left( -44 \right)}^{2}}-\left( 4\times 1\times 36 \right)}}{2\times 1} \\\ & =\dfrac{44\pm \sqrt{1936-144}}{2} \\\ & =\dfrac{44}{2}\pm \dfrac{\sqrt{1732}}{2} \\\ & =22\pm 8\sqrt{7} \\\ \end{aligned}$ So, we get $\sin 2x=22+8\sqrt{7}$ and $\sin 2x=22-8\sqrt{7}$. If we consider, $\sin 2x=22+8\sqrt{7}$, it will be rejected as $\left( -1\le \sin 2x\le 1 \right)$. So, we will consider $\sin 2x=22-8\sqrt{7}$, and we have been given in the question that $\sin 2x=a-b\sqrt{c}$, so on comparing both the equations, we will get, a = 22, b = 8 and c = 7 And we have been asked to find the value of the expression, $a-b+2c$. So, on substituting the values of a, b and c, we will get, $\begin{aligned} & a-b+2c \\\ & =22-8+2\left( 7 \right) \\\ & =14+14 \\\ & =28 \\\ \end{aligned}$ Therefore, we get the value of the expression, $a-b+2c$ as 28. **So, the correct answer is “Option C”.** **Note:** In this question students make mistakes in the last step, where we had to reject sin 2x = 0, because it is not possible for cot x and cosec x. Also, this question has a lot of calculations, so the students must make sure to do the calculations properly. So, keep in mind that we have to check the answer again to avoid the chances of any errors.