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# Prove the following expression $\dfrac{\sin {{70}^{o}}}{\cos {{20}^{o}}}+\dfrac{\csc {{20}^{o}}}{\sec {{70}^{o}}}-2\cos {{70}^{o}}\csc {{20}^{o}}=0$ Verified
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Hint: We convert all trigonometric functions in the denominator into corresponding complementary trigonometric functions using the below formulae.

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

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
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