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# If trigonometric ratio is given in the form of cosine as $\cos \theta =\dfrac{12}{13}$, show that $\sin \theta \left( 1-\tan \theta \right)=\dfrac{35}{156}$.

Last updated date: 28th Mar 2023
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Hint: First of all take LHS and convert the whole expression in terms of $\sin \theta$ and $\cos \theta$. Then use ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ to find the value of $\sin \theta$ by putting $\cos \theta =\dfrac{12}{13}$. Then put the values of $\sin \theta$ and $\cos \theta$ in LHS which is already in terms of $\sin \theta$ and $\cos \theta$ to find the required value.

We are given that $\cos \theta =\dfrac{12}{13}$. In this question, we have to show that $\sin \theta \left( 1-\tan \theta \right)=\dfrac{35}{156}$.
First of all, let us consider the left-hand side (LHS) of the above equation as,
$A=\sin \theta \left( 1-\tan \theta \right)$
We know that, $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$. By applying this in the above expression, we get,
$A=\sin \theta \left( 1-\dfrac{\sin \theta }{\cos \theta } \right)$
By simplifying the above expression, we get,
$A=\sin \theta \left( \dfrac{\cos \theta -\sin \theta }{\cos \theta } \right)$
By taking $\sin \theta$ inside the bracket in the above expression, we get,
$A=\dfrac{\sin \theta \cos \theta -{{\sin }^{2}}\theta }{\cos \theta }....\left( i \right)$
Now, we know that,
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1....\left( ii \right)$
Also, we are given that $\cos \theta =\dfrac{12}{13}$.
By putting the value of $\cos \theta$ in equation (ii), we get,
${{\sin }^{2}}\theta +{{\left( \dfrac{12}{13} \right)}^{2}}=1$
By simplifying the above equation, we get,
${{\sin }^{2}}\theta =1-\dfrac{144}{169}$
Or, ${{\sin }^{2}}\theta =\dfrac{169-144}{169}$
${{\sin }^{2}}\theta =\dfrac{25}{169}$
By taking square root on both the sides in the above equation, we get,
$\sin \theta =\sqrt{\dfrac{25}{169}}$
As we know ${{5}^{2}}=25$, therefore we get $\sqrt{25}=5$. Also, we know that ${{13}^{2}}=169$, therefore we get $\sqrt{169}=13$.
By putting the values of $\sqrt{25}$ and $\sqrt{169}$, we get,
$\Rightarrow \sin \theta =\dfrac{5}{13}$
Now, we put the value of $\cos \theta =\dfrac{12}{13}$ and $\sin \theta =\dfrac{5}{13}$ in equation (i), we get,
$A=\dfrac{\left( \cos \theta \right)\left( \sin \theta \right)-\left( {{\sin }^{2}}\theta \right)}{\cos \theta }$
$\Rightarrow A=\dfrac{\left( \dfrac{12}{13} \right)\left( \dfrac{5}{13} \right)-{{\left( \dfrac{5}{13} \right)}^{2}}}{\dfrac{12}{13}}$
By simplifying the above equation, we get,
$\Rightarrow A=\dfrac{\dfrac{60}{169}-\dfrac{25}{169}}{\dfrac{12}{13}}$
$A=\dfrac{\left( 60-25 \right)}{169}\times \dfrac{13}{12}$
Or, $A=\dfrac{35}{13\times 13}\times \dfrac{13}{12}$
By canceling the like terms, we get,
$A=\dfrac{35}{13\times 12}$
Or, $A=\dfrac{35}{156}\left[ \text{RHS} \right]$
Hence, we get LHS = RHS.
Therefore, we have shown that $\sin \theta \left( 1-\tan \theta \right)=\dfrac{35}{156}$ for $\cos \theta =\dfrac{12}{13}$.

Note: Here students must note that it is always better and reliable to convert the whole expression in terms of $\sin \theta$ and $\cos \theta$. Also, while using the formula ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ , students often forget to take the square root at the end to find the values of $\sin \theta$ and $\cos \theta$ and get confused by putting the values of ${{\sin }^{2}}\theta$ and ${{\cos }^{2}}\theta$ in place of $\sin \theta$ and $\cos \theta$. So, this mistake must be avoided.