Question

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

Answer 1

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.