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Question: If cos x = tan y, cot y = tan z and cot z = tan x, then sin x is equal to: (a) \[\dfrac{\sqrt{5}+1...

If cos x = tan y, cot y = tan z and cot z = tan x, then sin x is equal to:
(a) 5+14\dfrac{\sqrt{5}+1}{4}
(b) 5−14\dfrac{\sqrt{5}-1}{4}
(c) 5+12\dfrac{\sqrt{5}+1}{2}
(d) 5−12\dfrac{\sqrt{5}-1}{2}

Explanation

Solution

Hint: In this question, take any one of the given equations and try to substitute other equations by manipulating the first equation using various identities and ratios of trigonometry.

Complete step-by-step answer:
It is given that,
cos⁡x=tan⁡y....(i)\cos x=\tan y....\left( i \right)
cot⁡y=tan⁡z....(ii)\cot y=\tan z....\left( ii \right)
cot⁡z=tan⁡x....(iii)\cot z=\tan x....\left( iii \right)
Let us first solve cos x = tan y and try to simplify it.
We know by trigonometric ratios that, tan⁡y=1cot⁡y\tan y=\dfrac{1}{\cot y}
Substituting this value of tan y in the above equation we get,
cos⁡x=1cot⁡y\cos x=\dfrac{1}{\cot y}
Now we are given that tan z = cos y in equation (ii). Substituting this value of tan z in the above expression we get,
cos⁡x=1tan⁡z....(from(ii))\cos x=\dfrac{1}{\tan z}....\left( from\left( ii \right) \right)
By using the trigonometric ratio tan⁡z=1cot⁡z,\tan z=\dfrac{1}{\cot z},we get,
cos⁡x=cot⁡z\cos x=\cot z
Also, given equation (iii) is cot z = tan x…..(from (iii))
Substituting the value of cot z from equation (iii) in the above expression, we get,
cos⁡x=tan⁡x\cos x=\tan x
By using the trigonometric ratio, tan⁡x=sin⁡xcos⁡x,\tan x=\dfrac{\sin x}{\cos x}, we get,
cos⁡x=sin⁡xcos⁡x\cos x=\dfrac{\sin x}{\cos x}
Cross multiplying the above equation we get,
⇒cos⁡2x=sin⁡x\Rightarrow {{\cos }^{2}}x=\sin x
We have, sin⁡2x+cos⁡2x=1,{{\sin }^{2}}x+{{\cos }^{2}}x=1,we can write,
⇒cos⁡2x=1−sin⁡2x\Rightarrow {{\cos }^{2}}x=1-{{\sin }^{2}}x
Substituting this in the above equation, we get,
1−sin⁡2x=sin⁡x1-{{\sin }^{2}}x=\sin x
⇒sin⁡2x+sin⁡x−1=0\Rightarrow {{\sin }^{2}}x+\sin x-1=0
This is a quadratic equation. Therefore, by using the quadratic formula, we get,
sin⁡x=−b±b2−4ac2a\sin x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
Here, a = 1, b = 1 and c = – 1.
sin⁡x=−1±12−4(1)(−1)2(1)\sin x=\dfrac{-1\pm \sqrt{{{1}^{2}}-4\left( 1 \right)\left( -1 \right)}}{2\left( 1 \right)}
⇒sin⁡x=−1±1+42\Rightarrow \sin x=\dfrac{-1\pm \sqrt{1+4}}{2}
⇒sin⁡x=−1±52\Rightarrow \sin x=\dfrac{-1\pm \sqrt{5}}{2}
sin⁡x=−1+52 or sin⁡x=−1−52\sin x=\dfrac{-1+\sqrt{5}}{2}\text{ or }\sin x=\dfrac{-1-\sqrt{5}}{2}
Hence, −1−52<−1\dfrac{-1-\sqrt{5}}{2}<-1 and the value of sin x should lie between 0 and 1.
Therefore, ignoring −1−52,\dfrac{-1-\sqrt{5}}{2}, we are left with −1+52.\dfrac{-1+\sqrt{5}}{2}.
sin⁡x=−1+52\sin x=\dfrac{-1+\sqrt{5}}{2}
On rearranging, we get,
sin⁡x=5−12\sin x=\dfrac{\sqrt{5}-1}{2}
Hence, option (d) is the right answer.

Note: For these types of questions to be solved very easily, you should remember all the trigonometric formulas, those are – reciprocal identities, Pythagorean identities, sum and difference identities, double angle identities, and half-angle identities. Here, by looking at the options, you can guess that the quadratic formula has been used. Therefore, to get the final answer, you must try to simplify the given equation into a quadratic equation by using any of the above mentioned formulas.