Solveeit Logo
Login

Question

Question: If we have trigonometric function as \[\cos x=\tan y,\,\,\cot y=\tan z,\,\,\cot z=\tan x\] then \[\s...

If we have trigonometric function as cos⁡x=tan⁡y,  cot⁡y=tan⁡z,  cot⁡z=tan⁡x\cos x=\tan y,\,\,\cot y=\tan z,\,\,\cot z=\tan x then sin⁡x=\sin x=
A) sin⁡x=sin⁡y=sin⁡z=sin⁡(24∘)\sin x=\sin y=\sin z=\sin ({{24}^{\circ }})
B) sin⁡x=sin⁡y=sin⁡z=sin⁡(18∘)\sin x=\sin y=\sin z=\sin ({{18}^{\circ }})
C) sin⁡x=sin⁡y=sin⁡z=sin⁡(36∘)\sin x=\sin y=\sin z=\sin ({{36}^{\circ }})
D) sin⁡x=sin⁡y=sin⁡z=sin⁡(44∘)\sin x=\sin y=\sin z=\sin ({{44}^{\circ }})

Explanation

Solution

In this particular problem, first of all we have three equations. We have to multiply three equations and we get the answer to the term of sin⁡x\sin x in the form of a quadratic equation. By using the general formula for finding the roots we will get the value of sin⁡x\sin x. So, in this way we have an approach to solve this type of problem.

Complete step-by-step solution:
In this type of question we have three equation that is
cos⁡x=tan⁡y−−−(1)\cos x=\tan y---(1)
tan⁡z=cot⁡y−−−(2)\tan z=\cot y---(2)
cot⁡z=tan⁡x−−−(3)\cot z=\tan x---(3)
Multiply equation (1), equation (2) and equation (3) we get:
cos⁡xtan⁡zcot⁡z=tan⁡ycot⁡ytan⁡x\cos x\tan z\cot z=\tan y\cot y\tan x
As we know that tan⁡z=sin⁡zcos⁡z\tan z=\dfrac{\sin z}{\cos z}and cot⁡z=cos⁡zsin⁡z\cot z=\dfrac{\cos z}{\sin z}substitute in above equation we get:
cos⁡x×sin⁡zcos⁡z×cos⁡zsin⁡z=sin⁡ycos⁡y×cos⁡ysin⁡y×tan⁡x\cos x\times \dfrac{\sin z}{\cos z}\times \dfrac{\cos z}{\sin z}=\dfrac{\sin y}{\cos y}\times \dfrac{\cos y}{\sin y}\times \tan x
After simplification we get:
cos⁡x×1=1×tan⁡x\cos x\times 1=1\times \tan x
After here also we have to apply the trigonometry formula that is tan⁡z=sin⁡zcos⁡z\tan z=\dfrac{\sin z}{\cos z}
cos⁡x×1=1×sin⁡xcos⁡x\cos x\times 1=1\times \dfrac{\sin x}{\cos x}
After rearranging the term we get:
cos⁡2x=sin⁡x−−−(4){{\cos }^{2}}x=\sin x---(4)
Apply the trigonometry identity property that is sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 by arranging the term we get the value of cos⁡2x{{\cos }^{2}}x that is cos⁡2x=1−sin⁡2x{{\cos }^{2}}x=1-{{\sin }^{2}}x
Substitute this formula in equation (4) and write in the term of sin⁡x\sin x
Because in question we have to find the value of sin⁡x\sin x that we have to make the equation (4) in the term of sin⁡x\sin x only.
1−sin⁡2x=sin⁡x1-{{\sin }^{2}}x=\sin x
After simplifying and rearranging the term we get in the form of a quadratic form.
−sin⁡2x−sin⁡x+1=0-{{\sin }^{2}}x-\sin x+1=0
After multiplying -1 we get:
sin⁡2x+sin⁡x−1=0{{\sin }^{2}}x+\sin x-1=0
sin⁡x=−b±b2−4ac2a\sin x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
Here, b=1,  a=1,  c=−1b=1,\,\,a=1,\,\,c=-1 substitute these values in the above formula
sin⁡x=−1±(1)2−4(−1)2(1)\sin x=\dfrac{-1\pm \sqrt{{{(1)}^{2}}-4(-1)}}{2(1)}
After simplifying further we get:
sin⁡x=−1±1+42\sin x=\dfrac{-1\pm \sqrt{1+4}}{2}
Further solving we get:
sin⁡x=−1±52\sin x=\dfrac{-1\pm \sqrt{5}}{2}
Now, sin⁡(18∘)\sin ({{18}^{\circ }}) is positive, as 18∘{{18}^{\circ }} lies in the first quadrant.
sin⁡(18∘)=sin⁡A=sin⁡x=−1±52\sin ({{18}^{\circ }})=\sin A=\sin x=\dfrac{-1\pm \sqrt{5}}{2}
Hence sin⁡x=sin⁡y=sin⁡z=sin⁡(18∘)\sin x=\sin y=\sin z=\sin ({{18}^{\circ }})
So, the correct option is “option 2”.

Note: In this particular problem we always keep in mind that we need to form an equation in the term sin⁡x\sin x because it is asked to find sin⁡x\sin x. Another important thing is that you don't make silly mistakes while applying the formula and substitute the formula to get the desired values. Always keep in mind that for such types of problems, approach to problems should be good. So, the above solution is preferred for such types of problems.