Question

# (i) If we have a trigonometric equation $\cos 3A = \sin (A - {34^0}),$ where A is an acute angle, find the value of A.(ii) Prove the following identity, where the angles involved are acute angles for which the expression so defined is.$\dfrac{{1 + {{\cot }^2}A}}{{1 + {{\tan }^2}A}} = {\left( {\dfrac{{1 - \cot A}}{{1 - \tan A}}} \right)^2}$

Hint- In this question, simply use properties of acute angle i.e. for acute angle $\theta ,\cos \theta = \sin ({90^0} - \theta )$ and equate both sides to get the answer. For the second part, use basic trigonometric formulas and simplify both sides of the equation separately to check the result of L.H.S to R.H.S.

Complete step-by-step solution -
(i) Given that: $\cos 3A = \sin (A - {34^0})$ ---(a)
where A is acute angle $\theta ,\cos \theta = \sin ({90^0} - \theta )$ --(b)
From equation (a) we get
$\Rightarrow \cos 3A = \sin (A - {34^0})$
Using (b) we get
$\Rightarrow \sin ({90^0} - 3A) = \sin (A - {34^0})$
$\Rightarrow {90^0} - 3A = A - {34^0}$
$\Rightarrow {90^0} + {34^0} = A + 3A$
$\Rightarrow {124^0} = 4A$
$\Rightarrow \dfrac{{{{124}^0}}}{4} = A$
or, $A = {31^0}$
Hence, the value of $A = {31^0}$
$\therefore A = {31^0}$

(ii) To prove $\dfrac{{1 + {{\cot }^2}A}}{{1 + {{\tan }^2}A}} = {\left( {\dfrac{{1 - \cot A}}{{1 - \tan A}}} \right)^2}$ , we will just solve for the L.H.S and then equate that to the R.H.S.
L.H.S $= \dfrac{{1 + {{\cot }^2}A}}{{1 + {{\tan }^2}A}}$
Now, substitute $\cot A = \dfrac{{\cos A}}{{\sin A}}$ and $\tan A = \dfrac{{\sin A}}{{\cos A}}$
$= \dfrac{{1 + \left( {\dfrac{{{{\cos }^2}A}}{{{{\sin }^2}A}}} \right)}}{{1 + \left( {\dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}}} \right)}} = \dfrac{{\dfrac{{{{\sin }^2}A + {{\cos }^2}A}}{{{{\sin }^2}A}}}}{{\dfrac{{{{\cos }^2}A + {{\sin }^2}A}}{{{{\cos }^2}A}}}}$
Now, we know that ${\sin ^2}A + {\cos ^2}A = 1$
$= \dfrac{{\dfrac{1}{{{{\sin }^2}A}}}}{{\dfrac{1}{{{{\cos }^2}A}}}}$
Now, multiply and divide by $1 - 2\sin A\cos A$
$= \dfrac{{\dfrac{{1 - 2\sin A\cos A}}{{{{\sin }^2}A}}}}{{\dfrac{{1 - 2\sin A\cos A}}{{{{\cos }^2}A}}}}$
Split the denominator fraction of ${\sin ^2}A,{\cos ^2}A$
$= \dfrac{{\dfrac{1}{{{{\sin }^2}A}} - \dfrac{{2\sin A\cos A}}{{{{\sin }^2}A}}}}{{\dfrac{1}{{{{\cos }^2}A}} - \dfrac{{2\sin A\cos A}}{{{{\cos }^2}A}}}}$
Now, we know that $\dfrac{1}{{\sin A}} = \cos ecA,\dfrac{1}{{\cos A}} = \sec A$
$= \dfrac{{\cos e{c^2}A - \dfrac{{2\cos A}}{{\sin A}}}}{{{{\sec }^2}A - \dfrac{{2\sin A}}{{\cos A}}}}$
We know that $\cot A = \dfrac{{\cos A}}{{\sin A}}$ and $\tan A = \dfrac{{\sin A}}{{\cos A}}$
$= \dfrac{{\cos e{c^2}A - 2\cot A}}{{{{\sec }^2}A - 2\tan A}}$
Now, we know that $\cos e{c^2}A = 1 + {\cot ^2}A$ , $se{c^2}A = 1 + {\tan ^2}A$
$= \dfrac{{1 + {{\cot }^2}A - 2\cot A}}{{1 + {{\tan }^2}A - 2\tan A}} = \dfrac{{{1^2} + {{\cot }^2}A - 2\cot A}}{{{1^2} + {{\tan }^2}A - 2\tan A}}$
We know that, ${a^2} + {b^2} - 2ab = {(a - b)^2}$
$= \dfrac{{{{(1 - \cot A)}^2}}}{{{{(1 - \tan A)}^2}}} = {\left( {\dfrac{{1 - \cot A}}{{1 - \tan A}}} \right)^2} = R.H.S$
Now, L.H.S = R.H.S
Hence proved that $\dfrac{{1 + {{\cot }^2}A}}{{1 + {{\tan }^2}A}} = {\left( {\dfrac{{1 - \cot A}}{{1 - \tan A}}} \right)^2}$

Note- In such types of question, we just have to keep in mind of the trigonometric identities to simplify the equations using identities like $\cos \theta = \sin ({90^0} - \theta )$ , $\cot A = \dfrac{{\cos A}}{{\sin A}}$ , $\tan A = \dfrac{{\sin A}}{{\cos A}}$ , ${\sin ^2}A + {\cos ^2}A = 1$ and $\dfrac{1}{{\sin A}} = \cos ecA,\dfrac{1}{{\cos A}} = \sec A$ . Also, keep in mind of the algebraic identities like ${a^2} + {b^2} - 2ab = {(a - b)^2}$ .