Answer

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Hint: Relate \[\tan \theta \]and \[\cot \theta \] by changing their angles

(complementary to each other).

Here, we have given condition for calculation of angle \[A\] is

\[\tan 2A=\cot \left( A-{{18}^{o}} \right).....\left( i \right)\]

As we know that \[\tan \theta \]and \[\cot \theta \], both are reciprocal functions and can be

related in the following way: \[\tan \left( 90-\theta \right)=\cot \theta .....\left( ii \right)\]

Now, we can use property \[\left( ii \right)\]with the equation \[\left( i \right)\] to replace

\[\cot \left( A-{{18}^{o}} \right)\]

Let, \[A-{{18}^{o}}=\theta .....\left( iii \right)\]

So, the equation \[\left( i \right)\]becomes

\[\tan 2A=\cot \theta \]

Now, by using the equation \[\left( i \right)\], we can put \[\cot \theta =\tan \left( 90-\theta

\right)\] in the following manner:

\[\tan 2A=\cot \theta \]

\[\tan 2A=\tan \left( 90-\theta \right).....\left( iv \right)\]

As we know that if \[\tan {{\theta }_{1}}=\tan {{\theta }_{2}}\], then \[{{\theta }_{1}}\] and

\[{{\theta }_{2}}\] must be equal to get the same tangent of angles.

Hence, from the equation \[\left( iv \right)\], we can write

\[2A=90-\theta \]

Now, we already have \[\theta =A-18\] from the equation \[\left( iii \right)\].

Therefore, \[2A=90-\left( A-18 \right)\]

\[2A=90-A+18\]

\[3A=108\]

\[A={{36}^{o}}\]

As we already have given that \[A\] is an acute angle, so \[A={{36}^{o}}\] is correct.

Now, from \[\tan {{\theta }_{1}}=\tan {{\theta }_{2}}\], we can get infinite solutions by writing

the general solution of it in the following way:

If \[\tan {{\theta }_{1}}=\tan {{\theta }_{2}}\]

\[{{\theta }_{1}}=n\pi +{{\theta }_{2}}\] where \[n\in Z\left( n=0,1,2,3.... \right)\]

Hence, we can calculate more numbers of angles \[A\] but we need only an acute angle as

mentioned in the question. So, we need to find only the principle solution of \[\tan {{\theta

}_{1}}=\tan {{\theta }_{2}}\]or

\[\tan 2A=\tan \left( 108-A \right)\]

\[2A=n\pi +108-A\]

\[3A=n\pi +108\]

\[A=\dfrac{n\pi }{3}+36\]; \[n=0,1,2,3....\]

For, \[n=0,A={{36}^{o}}\] and for any other \[n\], \[A\] will be more than \[{{90}^{o}}\]. Hence

that will not be the correct solution for the given question.

Note: We can calculate angle \[A\] by converting \[\tan 2A\]to \[\cot \] function by using

\[\tan \theta =\cot \left( 90-\theta \right)\]

We have \[\tan 2A=\cot \left( A-18 \right)\]

\[\cot \left( 90-2A \right)=\cot \left( A-18 \right)\]

\[90-2A=A-18\]

\[A={{36}^{o}}\]

We can make a mistake when writing the equation \[\tan \left( 90-\theta \right)=\cot \theta

\]. One can get confused in the conversion of one function to another. So, he/she must use

the quadrant rules as follows.

(1)

(2) If the conversion is done like \[\dfrac{n\pi }{2}+\theta \] type, then we need to change

\[\tan \theta \rightleftarrows \cot \theta \]

\[\sin \theta \rightleftarrows \cos \theta \]

\[\text{cosec}\theta \rightleftarrows \sec \theta \]

where \[n\] is an odd integer.

If the conversion is like \[\left( n\pi +\theta \right)\] type, then we need not convert any

function to others only taking care of signs should be involved.

Example:

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

\[\dfrac{\pi }{2}+\theta \]will lie in the second quadrant, so \[\sin \] will be positive.

(2) \[\sin \left( \dfrac{3\pi }{2}+\theta \right)=-\cos \theta \]

\[\dfrac{3\pi }{2}+\theta \]will lie in the \[{{4}^{th}}\] quadrant.

(3) \[\tan \left( 2\pi -\theta \right)=-\tan \theta \]

\[2\pi -\theta \]will lie in the \[{{4}^{th}}\] quadrant.

(complementary to each other).

Here, we have given condition for calculation of angle \[A\] is

\[\tan 2A=\cot \left( A-{{18}^{o}} \right).....\left( i \right)\]

As we know that \[\tan \theta \]and \[\cot \theta \], both are reciprocal functions and can be

related in the following way: \[\tan \left( 90-\theta \right)=\cot \theta .....\left( ii \right)\]

Now, we can use property \[\left( ii \right)\]with the equation \[\left( i \right)\] to replace

\[\cot \left( A-{{18}^{o}} \right)\]

Let, \[A-{{18}^{o}}=\theta .....\left( iii \right)\]

So, the equation \[\left( i \right)\]becomes

\[\tan 2A=\cot \theta \]

Now, by using the equation \[\left( i \right)\], we can put \[\cot \theta =\tan \left( 90-\theta

\right)\] in the following manner:

\[\tan 2A=\cot \theta \]

\[\tan 2A=\tan \left( 90-\theta \right).....\left( iv \right)\]

As we know that if \[\tan {{\theta }_{1}}=\tan {{\theta }_{2}}\], then \[{{\theta }_{1}}\] and

\[{{\theta }_{2}}\] must be equal to get the same tangent of angles.

Hence, from the equation \[\left( iv \right)\], we can write

\[2A=90-\theta \]

Now, we already have \[\theta =A-18\] from the equation \[\left( iii \right)\].

Therefore, \[2A=90-\left( A-18 \right)\]

\[2A=90-A+18\]

\[3A=108\]

\[A={{36}^{o}}\]

As we already have given that \[A\] is an acute angle, so \[A={{36}^{o}}\] is correct.

Now, from \[\tan {{\theta }_{1}}=\tan {{\theta }_{2}}\], we can get infinite solutions by writing

the general solution of it in the following way:

If \[\tan {{\theta }_{1}}=\tan {{\theta }_{2}}\]

\[{{\theta }_{1}}=n\pi +{{\theta }_{2}}\] where \[n\in Z\left( n=0,1,2,3.... \right)\]

Hence, we can calculate more numbers of angles \[A\] but we need only an acute angle as

mentioned in the question. So, we need to find only the principle solution of \[\tan {{\theta

}_{1}}=\tan {{\theta }_{2}}\]or

\[\tan 2A=\tan \left( 108-A \right)\]

\[2A=n\pi +108-A\]

\[3A=n\pi +108\]

\[A=\dfrac{n\pi }{3}+36\]; \[n=0,1,2,3....\]

For, \[n=0,A={{36}^{o}}\] and for any other \[n\], \[A\] will be more than \[{{90}^{o}}\]. Hence

that will not be the correct solution for the given question.

Note: We can calculate angle \[A\] by converting \[\tan 2A\]to \[\cot \] function by using

\[\tan \theta =\cot \left( 90-\theta \right)\]

We have \[\tan 2A=\cot \left( A-18 \right)\]

\[\cot \left( 90-2A \right)=\cot \left( A-18 \right)\]

\[90-2A=A-18\]

\[A={{36}^{o}}\]

We can make a mistake when writing the equation \[\tan \left( 90-\theta \right)=\cot \theta

\]. One can get confused in the conversion of one function to another. So, he/she must use

the quadrant rules as follows.

(1)

(2) If the conversion is done like \[\dfrac{n\pi }{2}+\theta \] type, then we need to change

\[\tan \theta \rightleftarrows \cot \theta \]

\[\sin \theta \rightleftarrows \cos \theta \]

\[\text{cosec}\theta \rightleftarrows \sec \theta \]

where \[n\] is an odd integer.

If the conversion is like \[\left( n\pi +\theta \right)\] type, then we need not convert any

function to others only taking care of signs should be involved.

Example:

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

\[\dfrac{\pi }{2}+\theta \]will lie in the second quadrant, so \[\sin \] will be positive.

(2) \[\sin \left( \dfrac{3\pi }{2}+\theta \right)=-\cos \theta \]

\[\dfrac{3\pi }{2}+\theta \]will lie in the \[{{4}^{th}}\] quadrant.

(3) \[\tan \left( 2\pi -\theta \right)=-\tan \theta \]

\[2\pi -\theta \]will lie in the \[{{4}^{th}}\] quadrant.

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