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If $\cos \theta =0.6$, show that: $5\sin \theta -3\tan \theta =0$

Last updated date: 08th Sep 2024
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Hint: In this question, from the given values of cos function by using the trigonometric identities we can find the values of the sin and tan functions ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ and $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$. Then on substituting the respective values in the given expression of the question we can calculate the left hand side value and the right hand side value. Then on comparing the values obtained, we get the result.

Now, from the given question we have

$\cos \theta =0.6\ \ \ \ \ ...(a)$

Now, by using the trigonometric identity which gives the relation between the function that are mentioned in the hint, we get the following substitute the value from the question and as well as from equation (a)

$\Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$

Now substitute the $\cos \theta$ value and find the $\sin \theta$ value in the above equation then we get

$\Rightarrow {{\sin }^{2}}\theta =1-{{\left( 0.6 \right)}^{2}}$

$\Rightarrow {{\sin }^{2}}\theta =1-0.36$

On simplifying we get as follows

$\Rightarrow {{\sin }^{2}}\theta =0.64$

Since the perfect square of $0.64$ is $0.8$

$\Rightarrow \sin \theta =0.8$

Therefore, the value of $\sin \theta =0.8$

Now, again from the hint, we know that we can get the tan function as follows

$\Rightarrow \tan \theta =\dfrac{\sin \theta }{\cos \theta }$

And now let us substitute the $\sin \theta$ and $\cos \theta$ values in $\tan \theta$ then we get as follows

$\Rightarrow \tan \theta =\dfrac{0.8}{0.6}$

On simplifying we get as follows

$\Rightarrow \tan \theta =\dfrac{4}{3}$

Now, from the given expression in the question, on substituting the values, we have

$5\sin \theta -3\tan \theta =0$

Let us first consider the left hand side and calculate its value

$L.H.S=5\sin \theta -3\tan \theta$

And now substitute the $\sin \theta$ and $\tan \theta$ values then we get as follows

$L.H.S=5\times 0.8-3\times \dfrac{4}{3}$

On simplifying the above calculation we get as follows

$L.H.S=4-4=0$

Thus, the value of right hand side is equal to left hand side

Therefore, it is verified that $5\sin \theta -3\tan \theta =0$

Hence proved.

Note: The other way of solving the above problem is as follows:

We have given $\cos \theta =0.6$ so writing the value of $\cos \theta$ in terms of fraction we get,

$\cos \theta =\dfrac{6}{10}=\dfrac{3}{5}$

Now, we are going to draw a right angled triangle with angle $\theta$ as follows:

The length of AB is calculated using Pythagoras theorem which is equal to:

${{\left( Hypotenuse \right)}^{2}}={{\left( Perpendicular \right)}^{2}}+{{\left( Base \right)}^{2}}$

Substituting hypotenuse as AC, base as BC and perpendicular as AB in the above equation we get,

${{\left( AC \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$

$\Rightarrow {{5}^{2}}={{\left( AB \right)}^{2}}+{{3}^{2}}$

$\Rightarrow 25-9={{\left( AB \right)}^{2}}$

$\Rightarrow 16={{\left( AB \right)}^{2}}$

Taking square root on both the sides we get,

$4=AB$

Now, we can find $\sin \theta \And \tan \theta$ by using trigonometric ratios.

$\sin \theta =\dfrac{Perpendicular}{Hypotenuse}$

Substituting perpendicular as AB and hypotenuse as AC in the above we get,

$\sin \theta =\dfrac{AB}{AC}$

$\Rightarrow \sin \theta =\dfrac{4}{5}$

$\tan \theta =\dfrac{Perpendicular}{Base}$

$\Rightarrow \tan \theta =\dfrac{AB}{BC}$

$\Rightarrow \tan \theta =\dfrac{4}{3}$

Now, substitute the values of $\sin \theta \And \tan \theta$ in $5\sin \theta -3\tan \theta =0$ we get,

$5\left( \dfrac{4}{5} \right)-3\left( \dfrac{4}{3} \right)=0$

$\Rightarrow 4-4=0$

$\Rightarrow 0=0$

As L.H.S is equal to R.H.S so we have proved the given equation.