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If \[\cos ecA - \cot A = \dfrac{3}{2}\]. Find the value of \[\cos A\]

Last updated date: 13th Jun 2024
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Hint: We use the property of trigonometry \[1 + {\cot ^2}x = \cos e{c^2}x\] and bring constant to one side. Apply the formula of \[{a^2} - {b^2} = (a - b)(a + b)\] and open the trigonometric term. Using two equations and substitution methods using the value of cosecant of the angle. Use the fact that cosecant is reciprocal of sine function and then use the formula \[{\sin ^2}x + {\cos ^2}x = 1\] to find the cosine of the angle.

Complete step-by-step solution:
We know \[\cos ecA - \cot A = \dfrac{3}{2}\].................… (1)
We know \[1 + {\cot ^2}A = \cos e{c^2}A\]
Shift constant value to one side of the equation
\[ \Rightarrow 1 = \cos e{c^2}A - {\cot ^2}A\].....................… (2)
Now use the identity \[{a^2} - {b^2} = (a - b)(a + b)\]to open the terms in RHS
\[ \Rightarrow \cos e{c^2}A - {\cot ^2}A = (\cos ecA - \cot A)(\cos ecA + \cot A)\]
Put the values in LHS using equation (1) and in RHS using equation (2)
\[ \Rightarrow 1 = \dfrac{3}{2}(\cos ecA + \cot A)\]
Multiply both sides by \[\dfrac{2}{3}\]
\[ \Rightarrow \dfrac{2}{3} = \dfrac{2}{3} \times \dfrac{3}{2}(\cos ecA + \cot A)\]
Cancel same terms in RHS
\[ \Rightarrow \dfrac{2}{3} = \cos ecA + \cot A\].................… (3)
From equation (1) \[\cos ecA - \dfrac{3}{2} = \cot A\]
Substitute this value in equation (3)
\[ \Rightarrow \dfrac{2}{3} = \cos ecA + \cos ecA - \dfrac{3}{2}\]
Shift all constants on one side of the equation
\[ \Rightarrow \dfrac{2}{3} + \dfrac{3}{2} = 2\cos ecA\]
Take LCM in LHS
\[ \Rightarrow \dfrac{{2 \times 2 + 3 \times 3}}{6} = 2\cos ecA\]
\[ \Rightarrow \dfrac{{4 + 9}}{6} = 2\cos ecA\]
\[ \Rightarrow \dfrac{{13}}{6} = 2\cos ecA\]
Divide both sides of the equation by 2
\[ \Rightarrow \dfrac{{13}}{{6 \times 2}} = \cos ecA\]
\[ \Rightarrow \dfrac{{13}}{{12}} = \cos ecA\]
Since we know \[\cos ecx = \dfrac{1}{{\sin x}}\]
\[ \Rightarrow \sin A = \dfrac{{12}}{{13}}\]................… (4)
Now we know \[{\sin ^2}x + {\cos ^2}x = 1\]
\[ \Rightarrow {\cos ^2}A = 1 - {\sin ^2}A\]
Put the value of sine from equation (4)
\[ \Rightarrow {\cos ^2}A = 1 - {\left( {\dfrac{{12}}{{13}}} \right)^2}\]
\[ \Rightarrow {\cos ^2}A = 1 - \left( {\dfrac{{144}}{{169}}} \right)\]
Take LCM in RHS of the equation
\[ \Rightarrow {\cos ^2}A = \dfrac{{169 - 144}}{{169}}\]
\[ \Rightarrow {\cos ^2}A = \dfrac{{25}}{{169}}\]
We can write \[25 = {5^2},169 = {13^2}\]in RHS
\[ \Rightarrow {\cos ^2}A = \dfrac{{{5^2}}}{{{{13}^2}}}\]
Take square root on both sides of the equation
\[ \Rightarrow \sqrt {{{\cos }^2}A} = \sqrt {\dfrac{{{5^2}}}{{{{13}^2}}}} \]
Cancel square root by square power on both sides of the equation
\[ \Rightarrow \cos A = \dfrac{5}{{13}}\]

\[\therefore \]The value of \[\cos A\] is \[\dfrac{5}{{13}}\]

Note: Students many times make mistake of using the value of cosecant from equation (1) and use it in substitution, this will give us the value of cotangent and then we will have to involve lot more steps to reach up to the value of cosecant. Students are advised to directly aim for the value of cosecant as it will give us the value of sine easily which helps to find cosine value.