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MVSAT Dec 2023

Prove the given trigonometric expression: \[\dfrac{\cos {{27}^{\circ }}+\sin {{27}^{\circ }}}{\cos {{27}^{\circ }}-\sin {{27}^{\circ }}}=\tan {{72}^{\circ }}\]

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Hint: To prove the given trigonometric expression, divide the numerator and denominator by \[\cos {{27}^{\circ }}\]. We will get an expression in the form of tangent function. Replace 1 in the fraction by \[\tan {{45}^{\circ }}\] and use the formula for tangent of sum of two angles to prove the given expression.

Complete step-by-step answer:
We have to prove the given trigonometric expression \[\dfrac{\cos {{27}^{\circ }}+\sin {{27}^{\circ }}}{\cos {{27}^{\circ }}-\sin {{27}^{\circ }}}=\tan {{72}^{\circ }}\]. We will use trigonometric properties to do so.
We will simplify the left hand side of the function, i.e., \[\dfrac{\cos {{27}^{\circ }}+\sin {{27}^{\circ }}}{\cos {{27}^{\circ }}-\sin {{27}^{\circ }}}\].
Divide the numerator and denominator of the given expression by \[\cos {{27}^{\circ }}\].
Thus, we have \[\dfrac{\cos {{27}^{\circ }}+\sin {{27}^{\circ }}}{\cos {{27}^{\circ }}-\sin {{27}^{\circ }}}=\dfrac{1+\dfrac{\sin {{27}^{\circ }}}{\cos {{27}^{\circ }}}}{1-\dfrac{\sin {{27}^{\circ }}}{\cos {{27}^{\circ }}}}\].
We know that \[\dfrac{\sin x}{\cos x}=\tan x\].
Thus, we have \[\dfrac{\cos {{27}^{\circ }}+\sin {{27}^{\circ }}}{\cos {{27}^{\circ }}-\sin {{27}^{\circ }}}=\dfrac{1+\dfrac{\sin {{27}^{\circ }}}{\cos {{27}^{\circ }}}}{1-\dfrac{\sin {{27}^{\circ }}}{\cos {{27}^{\circ }}}}=\dfrac{1+\tan {{27}^{\circ }}}{1-\tan {{27}^{\circ }}}\].
We know that \[\tan {{45}^{\circ }}=1\].
Replacing 1 by \[\tan {{45}^{\circ }}\] in the above expression, we have \[\dfrac{\cos {{27}^{\circ }}+\sin {{27}^{\circ }}}{\cos {{27}^{\circ }}-\sin {{27}^{\circ }}}=\dfrac{1+\tan {{27}^{\circ }}}{1-\tan {{27}^{\circ }}}=\dfrac{\tan {{45}^{\circ }}+\tan {{27}^{\circ }}}{1-\tan {{45}^{\circ }}\tan {{27}^{\circ }}}\].
We know the trigonometric formula for tangent of sum of two angles, which is \[\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}\].
Substituting \[x={{45}^{\circ }},y={{27}^{\circ }}\] in the above expression, we have \[\tan \left( {{45}^{\circ }}+{{27}^{\circ }} \right)=\tan \left( {{72}^{\circ }} \right)=\dfrac{\tan {{45}^{\circ }}+\tan {{27}^{\circ }}}{1-\tan {{45}^{\circ }}\tan {{27}^{\circ }}}\].
Thus, we have \[\dfrac{\cos {{27}^{\circ }}+\sin {{27}^{\circ }}}{\cos {{27}^{\circ }}-\sin {{27}^{\circ }}}=\dfrac{\tan {{45}^{\circ }}+\tan {{27}^{\circ }}}{1-\tan {{45}^{\circ }}\tan {{27}^{\circ }}}=\tan {{72}^{\circ }}\].
Hence, we have proved that \[\dfrac{\cos {{27}^{\circ }}+\sin {{27}^{\circ }}}{\cos {{27}^{\circ }}-\sin {{27}^{\circ }}}=\tan {{72}^{\circ }}\].

Note: Trigonometric functions are real functions which relate any angle of a right angled triangle to the ratios of any two of its sides. The widely used trigonometric functions are sine, cosine and tangent. However, we can also use their reciprocals, i.e., cosecant, secant and cotangent. We can use geometric definitions to express the value of these functions on various angles using unit circle (circle with radius 1). We also write these trigonometric functions as infinite series or as solutions to differential equations. Thus, allowing us to expand the domain of these functions from the real line to the complex plane. One should be careful while using the trigonometric identities and rearranging the terms to convert from one trigonometric function to the other one.