Question

If we have a trigonometric expression as $3\sin x+4\cos x=5$ , then $6\tan \dfrac{x}{2}-9{{\tan }^{2}}\dfrac{x}{2}=$
( a ) 0
( b ) 1
( c ) 3
( d ) 4

Answer 1

Hint: In question it is asked that, we If $3\sin x+4\cos x=5$ , then $6\tan \dfrac{x}{2}-9{{\tan }^{2}}\dfrac{x}{2}$ is equals to what value. So, to do so we will use identities and properties of trigonometric ratios such as $\tan A=\dfrac{\sin A}{\cos A}$ and $\sin A=\dfrac{1}{\cos ecA}$ so as to obtain the $3\tan A=4\sin A$ in terms of cot A and cosec A .

Complete step-by-step solution:
We know that $\sin A$, $\cos A$, $\tan A$, $\cot A$, $\sec A$, and $\cos ec A$ are trigonometric function, where A is the angle made by the hypotenuse with the base of the triangle.
Now, in question it is given that $3\sin x+4\cos x=5$ .
Now, also we know that sin x can be written in terms of $\tan \dfrac{x}{2}$ function that is equals to $\dfrac{2\tan \dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}}$ and $\cos x$ can be written in terms of $\tan \dfrac{x}{2}$ function that is equals to $\dfrac{1-{{\tan }^{2}}\dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}}$ .
so, we can write $3\sin x+4\cos x=5$ in terms of $\tan \dfrac{x}{2}$ by putting $\sin x=\dfrac{2\tan \dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}}$ and $\cos x=\dfrac{1-{{\tan }^{2}}\dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}}$
we get,
$3\left( \dfrac{2\tan \dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}} \right)+4\left( \dfrac{1-{{\tan }^{2}}\dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}} \right)=5$
Taking L.C.M, we get
\[\dfrac{3\left( 2\tan \dfrac{x}{2} \right)+4\left( 1-{{\tan }^{2}}\dfrac{x}{2} \right)}{1+{{\tan }^{2}}\dfrac{x}{2}}=5\]
Taking \[1+{{\tan }^{2}}\dfrac{x}{2}\] from the denominator of left hand side to numerator of right hand side, using cross multiplication, we get
\[3\left( 2\tan \dfrac{x}{2} \right)+4\left( 1-{{\tan }^{2}}\dfrac{x}{2} \right)=5\cdot \left( 1+{{\tan }^{2}}\dfrac{x}{2} \right)\]
Opening brackets on left side and right hand side, we get
\[6\tan \dfrac{x}{2}+4-4{{\tan }^{2}}\dfrac{x}{2}=5+5{{\tan }^{2}}\dfrac{x}{2}\]
Shifting all values of left hand side to right hand side we get,
\[9{{\tan }^{2}}\dfrac{x}{2}-6\tan \dfrac{x}{2}+1=0\]
Shifting 1 from left hand side to right hand side, we get
 \[9{{\tan }^{2}}\dfrac{x}{2}-6\tan \dfrac{x}{2}=-1\]
Taking-1 common from the terms of left hand side we get
\[-\left( 6\tan \dfrac{x}{2}-9{{\tan }^{2}}\dfrac{x}{2} \right)=-1\]
Cancelling negative sign from left hand side with negative sign on right hand side we get,
\[\left( 6\tan \dfrac{x}{2}-9{{\tan }^{2}}\dfrac{x}{2} \right)=1\]
Hence, option ( b ) is correct.

Note: One must know all trigonometric identities which are ${{\cos }^{2}}x+{{\sin }^{2}}x=1$, $1+{{\tan }^{2}}x={{\sec }^{2}}x$ and $1+{{\cot }^{2}}x=\cos e{{c}^{2}}x$, properties and relation between trigonometric functions such as $\tan A=\dfrac{\sin A}{\cos A}$, $\cot A=\dfrac{\cos A}{\sin A}$, $\sin A=\dfrac{1}{\cos ecA}$, $\cot A=\dfrac{1}{\tan A}$ . While solving the question always use the most appropriate substitution of trigonometric relation which directly leads to result. There may be calculation mistake in cross multiplication, so be careful while solving an expression.