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# Prove the given trigonometric expression $\left( 1+\sec {{20}^{\circ }}+\cot {{70}^{\circ }} \right)\left( 1-\csc {{20}^{\circ }}+\tan {{70}^{\circ }} \right)=2$.

Last updated date: 24th Mar 2023
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Hint: Expand the left hand side of the given expression and rearrange the terms. Use trigonometric identities of secant, cosecant, tangent and cotangent functions to simplify the left hand side of the given expression and thus, prove the given expression.

We have to prove that $\left( 1+\sec {{20}^{\circ }}+\cot {{70}^{\circ }} \right)\left( 1-\csc {{20}^{\circ }}+\tan {{70}^{\circ }} \right)=2$.
We will expand the left hand side of the given expression.
Thus, we have $\left( 1+\sec {{20}^{\circ }}+\cot {{70}^{\circ }} \right)\left( 1-\csc {{20}^{\circ }}+\tan {{70}^{\circ }} \right)=1+\sec {{20}^{\circ }}+\cot {{70}^{\circ }}-\csc {{20}^{\circ }}\cot {{70}^{\circ }}-\csc {{20}^{\circ }}\sec {{20}^{\circ }}-\csc {{20}^{\circ }}+\tan {{70}^{\circ }}+\tan {{70}^{\circ }}\sec {{20}^{\circ }}+\tan {{70}^{\circ }}\cot {{70}^{\circ }}$.
We know that $\tan \theta \cot \theta =1$. Thus, we have $\tan {{70}^{\circ }}\cot {{70}^{\circ }}=1$.
We can rewrite ${{70}^{\circ }}$ as ${{70}^{\circ }}={{90}^{\circ }}-{{20}^{\circ }}$.
Thus, we have $\cot \left( {{70}^{\circ }} \right)=\cot \left( {{90}^{\circ }}-{{20}^{\circ }} \right)=\tan \left( {{20}^{\circ }} \right)$.
Similarly, we have $\tan \left( {{70}^{\circ }} \right)=\tan \left( {{90}^{\circ }}-{{20}^{\circ }} \right)=\cot \left( {{20}^{\circ }} \right)$.
Substituting the above equations in the expansion of the formula, we have $\left( 1+\sec {{20}^{\circ }}+\cot {{70}^{\circ }} \right)\left( 1-\csc {{20}^{\circ }}+\tan {{70}^{\circ }} \right)=1+\sec {{20}^{\circ }}+\tan {{20}^{\circ }}-\csc {{20}^{\circ }}\tan {{20}^{\circ }}-\csc {{20}^{\circ }}\sec {{20}^{\circ }}-\csc {{20}^{\circ }}+\cot {{20}^{\circ }}+\cot {{20}^{\circ }}\sec {{20}^{\circ }}+1$.
We know that $\csc \theta \tan \theta =\dfrac{1}{\sin \theta }\times \dfrac{\sin \theta }{\cos \theta }=\dfrac{1}{\cos \theta }=\sec \theta$. Thus $\csc {{20}^{\circ }}\tan {{20}^{\circ }}=\sec {{20}^{\circ }}$.
Similarly, we have $\csc \theta \sec \theta =\dfrac{1}{\sin \theta \cos \theta }$.
We have $\cot \theta \sec \theta =\dfrac{1}{\cos \theta }\times \dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\sin \theta }=\csc \theta$. Thus $\cot {{20}^{\circ }}\sec {{20}^{\circ }}=\csc {{20}^{\circ }}$.
Substituting the above equations in the expansion of the expression, we have $\left( 1+\sec {{20}^{\circ }}+\cot {{70}^{\circ }} \right)\left( 1-\csc {{20}^{\circ }}+\tan {{70}^{\circ }} \right)=2+\sec {{20}^{\circ }}+\tan {{20}^{\circ }}-\sec {{20}^{\circ }}-\dfrac{1}{\sin {{20}^{\circ }}\cos {{20}^{\circ }}}-\csc {{20}^{\circ }}+\cot {{20}^{\circ }}+\csc {{20}^{\circ }}$.
Simplifying the above expression, we have $\left( 1+\sec {{20}^{\circ }}+\cot {{70}^{\circ }} \right)\left( 1-\csc {{20}^{\circ }}+\tan {{70}^{\circ }} \right)=2+\tan {{20}^{\circ }}+\cot {{20}^{\circ }}-\dfrac{1}{\sin {{20}^{\circ }}\cos {{20}^{\circ }}}$.
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta },\cot \theta =\dfrac{\cos \theta }{\sin \theta }$.
Thus, we have $\left( 1+\sec {{20}^{\circ }}+\cot {{70}^{\circ }} \right)\left( 1-\csc {{20}^{\circ }}+\tan {{70}^{\circ }} \right)=2+\dfrac{\sin {{20}^{\circ }}}{\cos {{20}^{\circ }}}+\dfrac{\cos {{20}^{\circ }}}{\sin {{20}^{\circ }}}-\dfrac{1}{\sin {{20}^{\circ }}\cos {{20}^{\circ }}}$.
Further simplifying the expression, we have $\left( 1+\sec {{20}^{\circ }}+\cot {{70}^{\circ }} \right)\left( 1-\csc {{20}^{\circ }}+\tan {{70}^{\circ }} \right)=2+\dfrac{{{\sin }^{2}}{{20}^{\circ }}+{{\cos }^{2}}{{20}^{\circ }}}{\sin {{20}^{\circ }}\cos {{20}^{\circ }}}-\dfrac{1}{\sin {{20}^{\circ }}\cos {{20}^{\circ }}}$.
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
Thus, we have $\left( 1+\sec {{20}^{\circ }}+\cot {{70}^{\circ }} \right)\left( 1-\csc {{20}^{\circ }}+\tan {{70}^{\circ }} \right)=2+\dfrac{1}{\sin {{20}^{\circ }}\cos {{20}^{\circ }}}-\dfrac{1}{\sin {{20}^{\circ }}\cos {{20}^{\circ }}}$.
Further simplifying the expression, we get $\left( 1+\sec {{20}^{\circ }}+\cot {{70}^{\circ }} \right)\left( 1-\csc {{20}^{\circ }}+\tan {{70}^{\circ }} \right)=2$.
Hence, we have proved that $\left( 1+\sec {{20}^{\circ }}+\cot {{70}^{\circ }} \right)\left( 1-\csc {{20}^{\circ }}+\tan {{70}^{\circ }} \right)=2$.

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.