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Question: How do you prove \( \sin \theta \cot \theta =\cos \theta \) ?...

How do you prove sin⁡θcot⁡θ=cos⁡θ\sin \theta \cot \theta =\cos \theta ?

Explanation

Solution

Hint : We first try to find the respective ratios of the multiplication sin⁡θcot⁡θ\sin \theta \cot \theta . We find their respective values according to the sides of a right-angle triangle. We use the relations sin⁡θ=heighthypotenuse\sin \theta =\dfrac{height}{hypotenuse} and cot⁡θ=baseheight\cot \theta =\dfrac{base}{height} to multiply them. Then from the final ration we find the solution.

Complete step-by-step answer :
The given trigonometric expression is the multiplication of two ratios sin⁡θ\sin \theta and cot⁡θ\cot \theta .
We have their respective values according to the sides of a right-angle triangle. We use those relations to find the value of sin⁡θcot⁡θ\sin \theta \cot \theta .
According to a right-angle triangle the value of sin⁡θ\sin \theta will be considered as the ratio of the length of the height to the hypotenuse with respect to a certain angle.
So, sin⁡θ=heighthypotenuse\sin \theta =\dfrac{height}{hypotenuse} .
And according to the same right-angle triangle the value of cot⁡θ\cot \theta will be considered as the ratio of the length of the base to the height with respect to the same angle.
So, cot⁡θ=baseheight\cot \theta =\dfrac{base}{height} .
The multiplied form of the term sin⁡θcot⁡θ\sin \theta \cot \theta gives sin⁡θcot⁡θ=heighthypotenuse×baseheight=basehypotenuse\sin \theta \cot \theta =\dfrac{height}{hypotenuse}\times \dfrac{base}{height}=\dfrac{base}{hypotenuse} .
The ratio of basehypotenuse\dfrac{base}{hypotenuse} is defined as basehypotenuse=cos⁡θ\dfrac{base}{hypotenuse}=\cos \theta .
Therefore, sin⁡θcot⁡θ=cos⁡θ\sin \theta \cot \theta =\cos \theta .
So, the correct answer is “ sin⁡θcot⁡θ=cos⁡θ\sin \theta \cot \theta =\cos \theta ”.

Note : We can also use the direct ratio relation to find the solution. We know that cot⁡θ\cot \theta can be broken into a ratio of two other trigonometric expressions which are cos⁡θ\cos \theta and sin⁡θ\sin \theta . We know that cot⁡θ=cos⁡θsin⁡θ\cot \theta =\dfrac{\cos \theta }{\sin \theta } . Now we multiply sin⁡θ\sin \theta on both sides of the equation and get

& \cot \theta \times \sin \theta=\dfrac{\cos \theta }{\sin \theta }\times \sin \theta \\\ & \Rightarrow \sin \theta \cot \theta =\cos \theta \\\ \end{aligned}$$. Thus, verified the relation $ \sin \theta \cot \theta =\cos \theta $ .