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Question: If we are given an equation as \((1 + \sin x)(1 + \sin y)(1 + \sin z) = (1 - \sin x)(1 - \sin y)(1 -...

If we are given an equation as (1+sin⁡x)(1+sin⁡y)(1+sin⁡z)=(1−sin⁡x)(1−sin⁡y)(1−sin⁡z)=k(1 + \sin x)(1 + \sin y)(1 + \sin z) = (1 - \sin x)(1 - \sin y)(1 - \sin z) = k, then kk has the value
A) ±cos⁡xcos⁡ycos⁡z \pm \cos x\cos y\cos z
B) ±sin⁡xsin⁡ysin⁡z \pm \sin x\sin y\sin z
C) ±3sin⁡xsin⁡ysin⁡z \pm 3\sin x\sin y\sin z
D) None of these

Explanation

Solution

Trigonometric functions are essentially the functions of a triangle's angles. Sine, cosine, tangent, cotangent, secant, and cosecant are the fundamental trigonometric functions. Out of these six functions, sine and cosine are the primary classification of the functions and the rest can be derived from these three functions only. A very important identity of the sine and cosine function is sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1.

Complete step-by-step solution:
We are given that,
(1+sin⁡x)(1+sin⁡y)(1+sin⁡z)=(1−sin⁡x)(1−sin⁡y)(1−sin⁡z)=k(1 + \sin x)(1 + \sin y)(1 + \sin z) = (1 - \sin x)(1 - \sin y)(1 - \sin z) = k
The goal is to find the value of kk.
From the above equality we see that
k=(1+sin⁡x)(1+sin⁡y)(1+sin⁡z)k = (1 + \sin x)(1 + \sin y)(1 + \sin z), ------------------(1)
So, if we determine the value of this, we can get the value of kk.
Take,
⇒(1+sin⁡x)(1+sin⁡y)(1+sin⁡z)=(1−sin⁡x)(1−sin⁡y)(1−sin⁡z)\Rightarrow (1 + \sin x)(1 + \sin y)(1 + \sin z) = (1 - \sin x)(1 - \sin y)(1 - \sin z)
Multiply the given equality by (1+sin⁡x)(1+sin⁡y)(1+sin⁡z)(1 + \sin x)(1 + \sin y)(1 + \sin z), to get
⇒[(1+sin⁡x)(1+sin⁡y)(1+sin⁡z)]2=[(1−sin⁡x)(1+sin⁡x)][(1−sin⁡y)(1+sin⁡y)][(1−sin⁡z)(1+sin⁡z)]\Rightarrow {[(1 + \sin x)(1 + \sin y)(1 + \sin z)]^2} = [(1 - \sin x)(1 + \sin x)][(1 - \sin y)(1 + \sin y)][(1 - \sin z)(1 + \sin z)]
Use the algebraic identity given by (a−b)(a+b)=a2−b2(a - b)(a + b) = {a^2} - {b^2}, to evaluate the right side of the above equation
[(1+sin⁡x)(1+sin⁡y)(1+sin⁡z)]2=(1−sin⁡2x)(1−sin⁡2y)(1−sin⁡2z){[(1 + \sin x)(1 + \sin y)(1 + \sin z)]^2} = (1 - {\sin ^2}x)(1 - {\sin ^2}y)(1 - {\sin ^2}z)-----------------(2)
Since, we know that for the trigonometric function sin⁡θ\sin \theta and cos⁡θ\cos \theta , we have the identity which is given by,
sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
It can be written as,
cos⁡2θ=1−sin⁡2θ{\cos ^2}\theta = 1 - {\sin ^2}\theta
Substitute 1−sin⁡2θ1 - {\sin ^2}\theta with cos⁡2θ{\cos ^2}\theta in the right side of the equality in equation (2), to get
⇒[(1+sin⁡x)(1+sin⁡y)(1+sin⁡z)]2=(cos⁡2x)(cos⁡2y)(cos⁡2z)\Rightarrow {[(1 + \sin x)(1 + \sin y)(1 + \sin z)]^2} = ({\cos ^2}x)({\cos ^2}y)({\cos ^2}z)
Write the right side of the above equation in the whole square term,
⇒[(1+sin⁡x)(1+sin⁡y)(1+sin⁡z)]2=[cos⁡xcos⁡ycos⁡z]2\Rightarrow {[(1 + \sin x)(1 + \sin y)(1 + \sin z)]^2} = {[\cos x\cos y\cos z]^2}
Substitute the value of (1+sin⁡x)(1+sin⁡y)(1+sin⁡z)(1 + \sin x)(1 + \sin y)(1 + \sin z) by kk, from equation (1) in the above equality,
k2=[cos⁡xcos⁡ycos⁡z]2{k^2} = {[\cos x\cos y\cos z]^2}
Take the square root of the given above equation's two sides.,
k=±[cos⁡xcos⁡ycos⁡z]2k = \pm \sqrt {{{[\cos x\cos y\cos z]}^2}}
k=±cos⁡xcos⁡ycos⁡zk = \pm \cos x\cos y\cos z
So, the value of kk is ±cos⁡xcos⁡ycos⁡z \pm \cos x\cos y\cos z.
Hence, option A ±cos⁡xcos⁡ycos⁡z \pm \cos x\cos y\cos z is the correct option.

Note: If the equality is given in three terms, say a=b=ca = b = c, then it can be written in the form of three equalities, which are given by a=ba = b, b=cb = c and a=ca = c. If we want to determine the value of any of a,ba,b or cc, any pair of the given three equalities can be chosen to solve for it.