Answer

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Hint: We convert all trigonometric functions in the denominator into corresponding complementary trigonometric functions using the below formulae.

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

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

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

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

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.

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

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

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

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

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.

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