Question

If \[\cos ecA = 2\] then obtain the value of \[\cot A + \dfrac{{\sin A}}{{1 + \cos A}}\].

Answer 1

Hint: We will use the trigonometric identity \[{\sin ^2}A + {\cos ^2}A = 1\]. Also we will use the trigonometric ratios \[\sin A = \dfrac{1}{{\cos ecA}}\] to get the value of sinA and and use the identity to get the value of cosA. Next we will use \[\tan A = \dfrac{{\sin A}}{{\cos A}}\] to get the value of tanA and from this we can get the value of cotA by using \[\cot A = \dfrac{1}{{\tan A}}\]. Lastly, substituting all these values we will get the final output.

Complete step by step answer:
Given that,
\[\cos ecA = 2\]
Since, we know that,
\[\sin A = \dfrac{1}{{\cos ecA}}\]
Substituting the value of coescA we will get,
\[ \Rightarrow \sin A = \dfrac{1}{2}\]
We need to find the value of \[\cot A + \dfrac{{\sin A}}{{1 + \cos A}}\].
First we will use the trigonometry identity,
i.e. \[{\sin ^2}A + {\cos ^2}A = 1\]
Substituting the value of sinA in the above identity, we will get,
\[ \Rightarrow {(\dfrac{1}{2})^2} + {\cos ^2}A = 1\]
Simplify this above equation, we will get,
\[ \Rightarrow \dfrac{1}{4} + {\cos ^2}A = 1\]
By using the transposition method, we will move the LHS term to RHS.
\[ \Rightarrow {\cos ^2}A = 1 - \dfrac{1}{4}\]
Taking LCM as\[4\], we will get,
\[
   \Rightarrow {\cos ^2}A = \dfrac{{4 - 1}}{4} \\
   \Rightarrow {\cos ^2}A = \dfrac{3}{4} \\
 \]
Taking square root on both the sides, we will get,
\[
   \Rightarrow \cos A = \sqrt {\dfrac{3}{4}} \\
   \Rightarrow \cos A = \dfrac{{\sqrt 3 }}{2} \\
    \\
 \]
Also, we know that from the trigonometric ratios,
\[\tan A = \dfrac{{\sin A}}{{\cos A}}\]
Substituting the values in the above equation, we get,
\[ \Rightarrow \tan A = \dfrac{{\dfrac{1}{2}}}{{\dfrac{{\sqrt 3 }}{2}}}\]
\[ \Rightarrow \tan A = \dfrac{1}{2} \times \dfrac{2}{{\sqrt 3 }}\]
\[ \Rightarrow \tan A = \dfrac{1}{{\sqrt 3 }}\]
We know that,
\[\cot A = \dfrac{1}{{\tan A}}\]
\[ \Rightarrow \cot A = \dfrac{1}{{\dfrac{1}{{\sqrt 3 }}}}\]
\[
   \Rightarrow \cot A = 1 \times \sqrt 3 \\
   \Rightarrow \cot A = \sqrt 3 \\
 \]
Now, let’s find the value of
\[\cot A + \dfrac{{\sin A}}{{1 + \cos A}}\]
Substituting the values in the above expression, we will get,
\[ = \sqrt 3 + \dfrac{{\dfrac{1}{2}}}{{1 + \dfrac{{\sqrt 3 }}{2}}}\]
Taking LCM as \[2\] in the denominator part, we will get,
\[ = \sqrt 3 + \dfrac{{\dfrac{1}{2}}}{{\dfrac{{2 + \sqrt 3 }}{2}}}\]
\[ = \sqrt 3 + [\dfrac{1}{2} \times \dfrac{2}{{2 + \sqrt 3 }}]\]
\[ = \sqrt 3 + \dfrac{1}{{2 + \sqrt 3 }}\]
On evaluating this and taking LCM as \[2 + \sqrt 3 \], we will get,
\[ = \dfrac{{\sqrt 3 (2 + \sqrt 3 ) + 1}}{{2 + \sqrt 3 }}\]
\[ = \dfrac{{2\sqrt 3 + \sqrt 3 (\sqrt 3 ) + 1}}{{2 + \sqrt 3 }}\]
\[ = \dfrac{{2\sqrt 3 + 3 + 1}}{{2 + \sqrt 3 }}\]
Rationalising the denominator by multiplying this term by \[2 - \sqrt 3 \], we will get,
\[ = \dfrac{{2\sqrt 3 + 4}}{{2 + \sqrt 3 }} \times \dfrac{{2 - \sqrt 3 }}{{2 - \sqrt 3 }}\]
\[ = \dfrac{{(2\sqrt 3 + 4)(2 - \sqrt 3 )}}{{(2 + \sqrt 3 )(2 - \sqrt 3 )}}\]
We know that, \[(a - b)(a + b) = {a^2} - {b^2}\] and applying this, we will get,
\[ = \dfrac{{4\sqrt 3 - 2(\sqrt 3 \times \sqrt 3 ) + 8 - 4\sqrt 3 }}{{{2^2} - {{(\sqrt 3 )}^2}}}\]
On evaluating this further, we will get,
\[ = \dfrac{{4\sqrt 3 - 2(3) + 8 - 4\sqrt 3 }}{{4 - 3}}\]
\[
   = \dfrac{{ - 6 + 8}}{1} \\
   = 2 \\
 \]
Therefore, when \[\cos ecA = 2\], the value of \[\cot A + \dfrac{{\sin A}}{{1 + \cos A}} = 2\].

Additional information:
Trigonometry helps us to understand the relation between the sides and angles of a right-angle triangle. Cosec, Sec and Cot are three of the six trigonometric ratios of a right-angled triangle. The basics of trigonometry define three primary functions which are sine, cosine and tangent. The trigonometric ratios of a triangle are also called the trigonometric functions. We can solve this by another way, i.e. by taking angle value (\[A = {30^ \circ }\]) according to the given requirement. The angles are either measured in radians or degrees.

Note:
Alternative approach:
Given that,
\[\cos ecA = 2\]
Since, we know that,
\[\sin A = \dfrac{1}{{\cos ecA}}\]
Substituting the value of coescA we will get,
\[ \Rightarrow \sin A = \dfrac{1}{2}\]
\[\therefore A = {30^ \circ }\]
We need to find the value of \[\cot A + \dfrac{{\sin A}}{{1 + \cos A}}\].
\[\cot {30^ \circ } + \dfrac{{\sin {{30}^ \circ }}}{{1 + \cos {{30}^ \circ }}}\]
Substituting the values of all,
i.e.\[\cot {30^ \circ } = \sqrt 3 \] and \[\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}\]
We solve this as shown above.
Hence, when given \[\cos ecA = 2\] then the value of \[\cot A + \dfrac{{\sin A}}{{1 + \cos A}} = 2\].