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Prove the following expression \[\dfrac{\sin {{70}^{o}}}{\cos {{20}^{o}}}+\dfrac{\csc {{20}^{o}}}{\sec {{70}^{o}}}-2\cos {{70}^{o}}\csc {{20}^{o}}=0\]

Answer
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363.9k+ views
Hint: We convert all trigonometric functions in the denominator into corresponding complementary trigonometric functions using the below formulae.

\[\cos \theta \text{ }=\sin \left( \dfrac{\pi }{2}-\theta \right)\]
\[\sin \theta =\cos \left( \dfrac{\pi }{2}-\theta \right)\]
\[\csc \theta =\sec \left( \dfrac{\pi }{2}-\theta \right)\]
\[\sin \theta =\cos \left( \dfrac{\pi }{2}-\theta \right)\]


Complete step-by-step answer:

We have to prove the question \[\dfrac{\sin {{70}^{o}}}{\cos {{20}^{o}}}+\dfrac{\csc {{20}^{o}}}{\sec {{70}^{o}}}-2\cos {{70}^{o}}\csc {{20}^{o}}=0\]
Now, we will take left hand side i.e. \[\dfrac{\sin {{70}^{o}}}{\cos {{20}^{o}}}+\dfrac{\csc {{20}^{o}}}{\sec {{70}^{o}}}-2\cos {{70}^{o}}\csc {{20}^{o}}\]
Let \[\dfrac{\sin {{70}^{o}}}{\cos {{20}^{o}}}+\dfrac{\csc {{20}^{o}}}{\sec {{70}^{o}}}-2\cos {{70}^{o}}\csc {{20}^{o}}\] be equation 1.
Now we will simplify the denominators of equation 1 in the following way.
First we will convert \[\cos 20{}^\circ \] to \[\sin 70{}^\circ \] using the formula \[\cos \theta =\sin \left( \dfrac{\pi }{2}-\theta \right)\] ,
after converting \[\cos 20{}^\circ \] to \[\sin 70{}^\circ \],equation 1 becomes \[\dfrac{\sin {{70}^{o}}}{\sin {{70}^{o}}}+\dfrac{\csc {{20}^{o}}}{\sec {{70}^{o}}}-2\cos {{70}^{o}}\csc {{20}^{o}}\]
Now we will cancel equal terms in numerator and denominator, after cancelling common terms in numerator and denominator, equation 1 becomes
\[\Rightarrow 1+\dfrac{\csc {{20}^{o}}}{\sec {{70}^{o}}}-2\cos {{70}^{o}}\csc {{20}^{o}}\] and let it be equation 2.
Now we will convert a \[\sec 70{}^\circ \] to \[\csc 20{}^\circ \] using the formula \[\csc \theta =\sec \left( \dfrac{\pi }{2}-\theta \right)\] ,after converting
\[\sec 70{}^\circ \] to \[\text{csc20 }\!\!{}^\circ\!\!\text{ }\] equation 2 becomes \[\dfrac{\text{sin7}{{\text{0}}^{\text{o}}}}{\sin {{70}^{\text{o}}}}\text{+}\dfrac{\text{csc2}{{\text{0}}^{\text{o}}}}{\text{csc2}{{\text{0}}^{\text{o}}}}\text{-2cos7}{{\text{0}}^{\text{o}}}\text{csc2}{{\text{0}}^{\text{o}}}\] ,
now we will cancel common terms in numerator and denominator, after cancelling common terms in numerator and denominator equation 2 becomes
\[\Rightarrow 1+1-2\cos {{70}^{o}}\csc {{20}^{o}}\]
 \[\Rightarrow 2-2\cos {{70}^{o}}\csc {{20}^{o}}\] and let it be equation 3.
We know that \[\csc\theta \] and \[\sin \theta \] are mutual reciprocals, that means $\csc \theta =\dfrac{1}{\sin \theta }$ or $\sin \theta =\dfrac{1}{\csc \theta }$
Now we will convert a$\csc 20{}^\circ $ to $\dfrac{1}{\sin 20{}^\circ }$ and equation 3 becomes
$2-2\dfrac{\cos 70{}^\circ }{\sin 20{}^\circ }$ let it be equation 4.
Now we will convert \[\sin 20{}^\circ \] to $\cos 70{}^\circ $ using the formula \[\sin \theta =\cos \left( \dfrac{\pi }{2}-\theta \right)\] ,after converting
\[\sin 20{}^\circ \] to $\cos 70{}^\circ $ equation 4 becomes $2-2\dfrac{\cos 70{}^\circ }{\cos 70{}^\circ }$ .
Now we cancel common terms in numerator and denominator, after cancelling common terms in numerator and denominator equation 4 becomes
$\Rightarrow 2-2\times 1$
\[\Rightarrow 0\]
=RHS
LHS=RHS
Hence, the following expression \[\text{ }\dfrac{\sin {{70}^{o}}}{\cos {{20}^{o}}}+\dfrac{\csc {{20}^{o}}}{\sec {{70}^{o}}}-2\cos {{70}^{o}}\csc {{20}^{o}}=0\] is proved.
So, \[\text{ }\dfrac{\text{sin7}{{\text{0}}^{\text{o}}}}{\text{cos2}{{\text{0}}^{\text{o}}}}\text{+}\dfrac{\text{csc2}{{\text{0}}^{\text{o}}}}{\text{sec7}{{\text{0}}^{\text{o}}}}\text{-2cos7}{{\text{0}}^{\text{o}}}\text{csc2}{{\text{0}}^{\text{o}}}\text{=0}\]

Note: Sometimes we convert trigonometric functions of non-standard angles to corresponding complementary trigonometric functions i.e. \[\cos \theta \] to \[\sin \left( \dfrac{\pi }{2}-\theta \right)\], such that this conversion will help us to simplify the trigonometric equations and then there will be a confusion in converting trigonometric functions into corresponding complementary trigonometric functions i.e. \[\cos \theta \] to \[\sin \left( \dfrac{\pi }{2}-\theta \right)\] so, be perfect with this type of basic formulae here.
Last updated date: 01st Oct 2023
Total views: 363.9k
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