Question

If the equation $\sec x+\tan x=2$ holds true then which of the following options are correct.
$\begin{align}
  & a)x\in \left( 2n\pi ,2n\pi +\dfrac{\pi }{2} \right) \\
 & b)x\in \left( (2n+1)\pi ,2n\pi +\dfrac{3\pi }{2} \right) \\
 & c)\cos 2x=\dfrac{7}{25} \\
 & d)\tan ({{45}^{\circ }}-x)=\dfrac{1}{7} \\
\end{align}$

Answer 1

Hint: we know the relation between $\tan x$and $\sec x$ which is ${{\tan }^{2}}\theta +1={{\sec }^{2}}\theta $ we will try to use this equation to find $\tan x-\sec x$ and then we will solve it with the given equation to find $\tan x$and $\sec x$. Hence we can find the required condition on x as well as find the value of $\cos 2x$ and $\tan (45-x)$ with the formula $\cos 2x=2{{\cos }^{2}}x-1$ and $\tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a\tan b}$ respectively.

Complete step by step answer:
Now we are given that
$\tan x+\sec x=2.............(1)$
Now we know the trigonometric identity which says ${{\tan }^{2}}x+1={{\sec }^{2}}x$
Rearranging the terms we get ${{\sec }^{2}}x-{{\tan }^{2}}x=1$
Now we know ${{a}^{2}}-{{b}^{2}}=(a-b)(a+b)$ using this equality in above equation we get
$(\sec x+\tan x)(\sec x-\tan x)=1$
But from equation (1) we have the value of $\tan x+\sec x=2$ hence we have
$2(\sec x-\tan x)=1$
Now dividing both sides by 2 we get
$\sec x-\tan x=\dfrac{1}{2}....................(2)$
Now let us add equation (1) and equation (2).
$\begin{align}
  & 2\sec x=2+\dfrac{1}{2} \\
 & \Rightarrow 2\sec x=\dfrac{4+1}{2} \\
 & \Rightarrow 2\sec x=\dfrac{5}{2} \\
 & \Rightarrow \sec x=\dfrac{5}{4} \\
\end{align}$
Now substituting the value of $\sec x$ in equation (1) we get
$\begin{align}
  & \dfrac{5}{4}+\tan x=2 \\
 & \Rightarrow \tan x=2-\dfrac{5}{4} \\
 & \Rightarrow \tan x=\dfrac{8-5}{4} \\
 & \Rightarrow \tan x=\dfrac{3}{4}...........................(3) \\
\end{align}$
Now we know that
$\begin{align}
  & \cos x=\dfrac{1}{\sec x} \\
 & \Rightarrow \cos x=\dfrac{4}{5}...................(4) \\
\end{align}$
Now $\tan x$ is nothing but $\dfrac{\sin x}{\cos x}$
Hence we have
 \[\begin{align}
  & \tan x=\dfrac{\sin x}{\cos x} \\
 & \Rightarrow \dfrac{3}{4}=\dfrac{\sin x}{\dfrac{4}{5}} \\
 & \Rightarrow \dfrac{3}{4}\times \dfrac{4}{5}=\sin x \\
 & \Rightarrow \sin x=\dfrac{3}{5}.....................(5) \\
\end{align}\]
Now from equation (3), equation (4) and equation (5) we have $\sin x,\cos x,\tan x > 0$
This Condition is true only for the first quadrant. Hence we know now that x lies in the first quadrant.
Now first quadrant means $x\in \left( 0,\dfrac{\pi }{2} \right)$
Hence in general we write it as
\[\begin{align}
  & x\in \left( 0+2n\pi ,\dfrac{\pi }{2}+2n\pi \right) \\
 &\Rightarrow x\in \left( 2n\pi ,\dfrac{\pi }{2}+2n\pi \right) \\
\end{align}\]
Hence $x\in (2n\pi ,2n\pi +\dfrac{\pi }{2})$.
Option a is correct option
Now $x\in \left( (2n+1)\pi ,2n\pi +\dfrac{3\pi }{2} \right)$ means that x belongs to third quadrant as for n = 0 we have $x\in \left( \pi ,\dfrac{3\pi }{2} \right)$ . But in the third quadrant sin and cos are negative. But from equation (4) and equation (5) we have sin and cos are positive
Hence, option b is incorrect
Now from equation (4) we have $\cos x=\dfrac{4}{3}$
We have a formula for $\cos 2x$ which is $\cos 2x=2{{\cos }^{2}}x-1$ applying this formula we get
$\begin{align}
  & \cos 2x=2{{\left( \dfrac{4}{3} \right)}^{2}}-1 \\
 & \cos 2x=2\left( \dfrac{16}{9} \right)-1 \\
 & =\dfrac{32}{9}-1 \\
 & =\dfrac{32-9}{9} \\
 & =\dfrac{23}{9} \\
\end{align}$
Hence option c is incorrect
Now let us check the value of $\tan \left( {{45}^{\circ }}-x \right)$
We know $\tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a\tan b}$ and $\tan {{45}^{\circ }}=1$
Hence we get $\tan \left( {{45}^{\circ }}-x \right)=\dfrac{\tan 45-\tan x}{1+\tan 45\tan x}$
$=\dfrac{1-\tan x}{1+\tan x}$
Now from equation (3) we get
$\begin{align}
  &\Rightarrow \tan \left( {{45}^{\circ }}-x \right)=\dfrac{1-\dfrac{3}{4}}{1+\dfrac{3}{4}} \\
 & \Rightarrow \tan \left( {{45}^{\circ }}-x \right)=\dfrac{\dfrac{4-3}{4}}{\dfrac{4+3}{4}} \\
 &\Rightarrow \tan \left( {{45}^{\circ }}-x \right)=\dfrac{1}{7} \\
\end{align}$
Hence option d is also correct.

So, the correct answer is “Option A and D”.

Note: note that the 1st quadrant consists of angle $\left( 0,\dfrac{\pi }{2} \right)$. But after 360° rotation we will arrive at the same angle hence all the multiples of 360° are added and we get 1st quadrant as $\left( 2n\pi ,2n\pi +\dfrac{\pi }{2} \right)$. Also $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$ and $\cos 2x=1-2{{\sin }^{2}}x$. Any of this formula can be used to find $\cos 2x$