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Question: If angle \(\theta \) is divided into two parts such that the tangent of one part is \(\lambda \) tim...

If angle θ\theta is divided into two parts such that the tangent of one part is λ\lambda times the tangent of the other part, and ϕ\phi is their difference then show that sin⁡θ=λ+1λ−1sin⁡ϕ\sin \theta =\dfrac{\lambda +1}{\lambda -1}\sin \phi .

Explanation

Solution

Hint: Try to simplify the left-hand side of the equation given in the question. Start by taking one part of the angle θ\theta be x and the other part be y. Now form the equations given in the question and use the formula of sin(A+B) to solve the left-hand side of the equation given in the question.

Complete step-by-step answer:
Before we start with the solution, let us draw a diagram for better visualisation.

According to the question:
x+y=θx+y=\theta
x−y=ϕx-y=\phi
Now starting with the left-hand side of the equation that is given in the question. If we put x+y=θx+y=\theta in the expression, we get:
sin⁡θ\sin \theta
=sin⁡(x+y)=\sin \left( x+y \right)
Now we will divide and multiply the expression by sin(x-y). On doing so, we get
sin⁡(x+y)sin⁡(x−y)sin⁡(x−y)\dfrac{\sin \left( x+y \right)\sin \left( x-y \right)}{\sin \left( x-y \right)}
Now we know that sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B and sin(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\text{sin}\left( A+B \right)=\sin A\cos B+\cos A\sin B . On using this in our expression, we get
sin⁡xcos⁡y+cos⁡xsin⁡ysin⁡xcos⁡y−cos⁡xsin⁡ysin(x−y)\dfrac{\sin x\cos y+\cos x\sin y}{\sin x\cos y-\cos x\sin y}sin\left( x-y \right)
Now we will divide both the numerator and the denominator of the expression by cosxcosy. On doing so, we get
sin⁡xcos⁡ycos⁡xcos⁡y+cos⁡xsin⁡ycos⁡xcos⁡ysin⁡xcos⁡ycos⁡xcos⁡y+cos⁡xsin⁡ycos⁡xcos⁡ysin(x−y)\dfrac{\dfrac{\sin x\cos y}{\cos x\cos y}+\dfrac{\cos x\sin y}{\cos x\cos y}}{\dfrac{\sin x\cos y}{\cos x\cos y}+\dfrac{\cos x\sin y}{\cos x\cos y}}sin\left( x-y \right)
We know that tan⁡A=sin⁡Acos⁡A\tan A=\dfrac{\sin A}{\cos A} .
∴sin⁡xcos⁡x+sin⁡ycos⁡ysin⁡xcos⁡x−sin⁡ycos⁡ysin(x−y)\therefore \dfrac{\dfrac{\sin x}{\cos x}+\dfrac{\sin y}{\cos y}}{\dfrac{\sin x}{\cos x}-\dfrac{\sin y}{\cos y}}sin\left( x-y \right)
=tan⁡x+tan⁡ytan⁡x−tan⁡ysin(x−y)=\dfrac{\tan x+\tan y}{\tan x-\tan y}sin\left( x-y \right)
From the question, we can also say that: tan⁡x=λtan⁡y\tan x=\lambda \tan y . So, substituting tanx in our expression, we get
=λtan⁡y+tan⁡yλtan⁡y−tan⁡ysin(x−y)=\dfrac{\lambda \tan y+\tan y}{\lambda \tan y-\tan y}sin\left( x-y \right)
=tan⁡y(λ+1)tan⁡y(λ−1)sin(x−y)=\dfrac{\tan y\left( \lambda +1 \right)}{\tan y\left( \lambda -1 \right)}sin\left( x-y \right)
=λ+1λ−1sin(x−y)=\dfrac{\lambda +1}{\lambda -1}sin\left( x-y \right)
Now substituting x-y as ϕ\phi , we get
=λ+1λ−1sinϕ=\dfrac{\lambda +1}{\lambda -1}sin\phi
The left-hand side of the equation given in the question is equal to the right-hand side of the equation. Hence, we can say that we have proved the equation given in the question.

Note: Be careful about the calculation and the signs while opening the brackets. The general mistake that a student can make is 1+x-(x-1)=1+x-x-1. Also, be careful about the signs in the formula of sin(A-B) and sin(A+B). Whenever you are dealing with an expression having cotangents, secant, and cosecant involved, it is better to convert it to an equivalent expression in terms of sine, cosine, and tangent, as most of the formulas we know are valid for sine, cosine, and tangents only.