Question

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

Answer 1

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).
$\begin{aligned} & 2\sec x=2+\dfrac{1}{2} \\\ & \Rightarrow 2\sec x=\dfrac{4+1}{2} \\\ & \Rightarrow 2\sec x=\dfrac{5}{2} \\\ & \Rightarrow \sec x=\dfrac{5}{4} \\\ \end{aligned}$
Now substituting the value of $\sec x$ in equation (1) we get
$\begin{aligned} & \dfrac{5}{4}+\tan x=2 \\\ & \Rightarrow \tan x=2-\dfrac{5}{4} \\\ & \Rightarrow \tan x=\dfrac{8-5}{4} \\\ & \Rightarrow \tan x=\dfrac{3}{4}...........................(3) \\\ \end{aligned}$
Now we know that
$\begin{aligned} & \cos x=\dfrac{1}{\sec x} \\\ & \Rightarrow \cos x=\dfrac{4}{5}...................(4) \\\ \end{aligned}$
Now $\tan x$ is nothing but $\dfrac{\sin x}{\cos x}$
Hence we have

& \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{aligned}$$ Now from equation (3), equation (4) and equation (5) we have $\sin x,\cos x,\tan x > 0$ This Condition is true only for the first quadrant. Hence we know now that x lies in the first quadrant. Now first quadrant means $x\in \left( 0,\dfrac{\pi }{2} \right)$ Hence in general we write it as $$\begin{aligned} & 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{aligned}$$ 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 $\begin{aligned} & \cos 2x=2{{\left( \dfrac{4}{3} \right)}^{2}}-1 \\\ & \cos 2x=2\left( \dfrac{16}{9} \right)-1 \\\ & =\dfrac{32}{9}-1 \\\ & =\dfrac{32-9}{9} \\\ & =\dfrac{23}{9} \\\ \end{aligned}$ 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 $\begin{aligned} &\Rightarrow \tan \left( {{45}^{\circ }}-x \right)=\dfrac{1-\dfrac{3}{4}}{1+\dfrac{3}{4}} \\\ & \Rightarrow \tan \left( {{45}^{\circ }}-x \right)=\dfrac{\dfrac{4-3}{4}}{\dfrac{4+3}{4}} \\\ &\Rightarrow \tan \left( {{45}^{\circ }}-x \right)=\dfrac{1}{7} \\\ \end{aligned}$ 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 $\cos 2x$