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Prove the following:
A) $ \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = 1 + \sec A\cos ecA $
B) $ \dfrac{{\cos ecA}}{{\left( {\cos ecA - 1} \right)}} + \dfrac{{\cos ecA}}{{\left( {\cos ecA + 1} \right)}} = 2{\sec ^2}A $

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Last updated date: 20th Jun 2024
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Answer
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Hint: Initially, define the L.H.S part and R.H.S part of the given expression. Then, consider L.H.S part and simplify it by using the trigonometric formulas along with algebraic identities. Then, check whether the L.H.S part is similar to the R.H.S part or not. If it is equal then, the given equation will get proved.

Complete step-by-step answer:
Given expression is:
\[ \Rightarrow \]$ \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = 1 + \sec A\cos ecA $----- (1)
From equation (1),
L.H.S = $ \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} $
R.H.S = $ 1 + \sec A\cos ecA $
Consider the left hand side of the given expression ‘L.H.S’,
\[ \Rightarrow \]$ \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} $----- (2)
Using trigonometric formula, we know that:
\[ \Rightarrow \]$ \tan A = \dfrac{{\sin A}}{{\cos A}} $----- (3)
\[ \Rightarrow \]$ \cot A = \dfrac{{\cos A}}{{\sin A}} $----- (4)
Substitute equation (3) and (4) into equation (2), and we get:
\[ \Rightarrow \]$ \dfrac{{\dfrac{{\sin A}}{{\cos A}}}}{{1 - \dfrac{{\cos A}}{{\sin A}}}} + \dfrac{{\dfrac{{\cos A}}{{\sin A}}}}{{1 - \dfrac{{\sin A}}{{\cos A}}}} $
\[ \Rightarrow \]$ \dfrac{{\dfrac{{\sin A}}{{\cos A}}}}{{\dfrac{{\sin A - \cos A}}{{\sin A}}}} + \dfrac{{\dfrac{{\cos A}}{{\sin A}}}}{{\dfrac{{co\operatorname{s} A - \sin A}}{{co\operatorname{s} A}}}} $
\[ \Rightarrow \]$ \dfrac{{{{\sin }^2}A}}{{\cos A(\sin A - \cos A)}} + \dfrac{{{{\cos }^2}A}}{{\sin A(\cos A - \sin A)}} $
\[ \Rightarrow \]$ \dfrac{{{{\sin }^2}A}}{{\cos A(\sin A - \cos A)}} - \dfrac{{{{\cos }^2}A}}{{\sin A(\sin A - \cos A)}} $
\[ \Rightarrow \]$ \dfrac{{{{\sin }^3}A}}{{\cos A\sin A(\sin A - \cos A)}} - \dfrac{{{{\cos }^3}A}}{{\sin A\cos A(\sin A - \cos A)}} $
\[ \Rightarrow \]$ \dfrac{{{{\sin }^3}A - {{\cos }^3}A}}{{\cos A\sin A(\sin A - \cos A)}} $ ----- (5)
Using Algebraic Identities, $ {a^3} - {b^3} = (a - b)({a^2} + ab + {b^2}) $ in equation (5), we get:
\[ \Rightarrow \]$ \dfrac{{(\sin A - \cos A)({{\sin }^2}A + \sin A\cos A + {{\cos }^2}A)}}{{\cos A\sin A(\sin A - \cos A)}} $
\[ \Rightarrow \]$ \dfrac{{({{\sin }^2}A + \sin A\cos A + {{\cos }^2}A)}}{{\cos A\sin A}} $----- (6)
Using Trigonometric Identities: $ {\sin ^2}A + {\cos ^2}A = 1 $ in equation (6), we get:
\[ \Rightarrow \]$ \dfrac{{(1 + \sin A\cos A)}}{{\cos A\sin A}} $ ----- (7)
Using Trigonometric formula: $ \dfrac{1}{{\cos A}} = \sec A, $ $ \dfrac{1}{{\sin A}} = \cos ecA $ in equation (7), we get:
\[ \Rightarrow \]$ \sec A\cos ecA + 1 $ = R.H.S ----- (8)
It is clear from equation (2) and (8) that
L.H.S = R.H.S
\[ \Rightarrow \]$ \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = 1 + \sec A\cos ecA $
Given expression is:
\[ \Rightarrow \] $ \dfrac{{\cos ecA}}{{\left( {\cos ecA - 1} \right)}} + \dfrac{{\cos ecA}}{{\left( {\cos ecA + 1} \right)}} = 2{\sec ^2}A $----- (1)
From equation (1),
L.H.S = $ \dfrac{{\cos ecA}}{{\left( {\cos ecA - 1} \right)}} + \dfrac{{\cos ecA}}{{\left( {\cos ecA + 1} \right)}} $
R.H.S = $ 2{\sec ^2}A $
Consider the left hand side of the given expression ‘L.H.S’,
\[ \Rightarrow \]$ \dfrac{{\cos ecA}}{{\left( {\cos ecA - 1} \right)}} + \dfrac{{\cos ecA}}{{\left( {\cos ecA + 1} \right)}} $----- (2)
Taking L.C.M of equation (2), and simplifying it, we get:
\[ \Rightarrow \]$ \dfrac{{\cos ecA\left( {\cos ecA + 1} \right) + \cos ecA\left( {\cos ecA - 1} \right)}}{{\left( {\cos ecA - 1} \right)\left( {\cos ecA + 1} \right)}} $
Using Algebraic Identities, $ {a^2} - {b^2} = (a - b)(a + b) $ in above expression, we get:
\[ \Rightarrow \]$ \dfrac{{2\cos e{c^2}A}}{{\cos e{c^2}A - 1}}$ ----- (3)
Using Trigonometric formula:
\[ \Rightarrow \]$ \dfrac{1}{{\sin A}} = \cos ecA $ and $\cos e{c^2}A - 1 = {\cot ^2} A$ in equation (3), we get:
\[ \Rightarrow \]$ \dfrac{{2\cos e{c^2}A}}{{{{\cot }^2}A}} $
\[ \Rightarrow \]\[ \dfrac{{\dfrac{2}{{{{\sin }^2}A}}}}{{\dfrac{{{{\cos }^2}A}}{{{{\sin }^2}A}}}} = \dfrac{2}{{{{\sin }^2}A}} \times \dfrac{{{{\sin }^2}A}}{{co{\operatorname{s} ^2}A}}\]
\[ \to \dfrac{2}{{co{\operatorname{s} ^2}A}} = 2{\sec ^2}A\] ----- (4)
It is clear from equation (2) and (4) that
L.H.S = R.H.S
\[ \Rightarrow \]$ \dfrac{{\cos ecA}}{{\left( {\cos ecA - 1} \right)}} + \dfrac{{\cos ecA}}{{\left( {\cos ecA + 1} \right)}} = 2{\sec ^2}A $

Note: In conclusion always try to convert any given trigonometric expression into basic ‘sin’ and ‘cos’ terms and using algebraic identities formula, it can be easily solved. Whenever the ‘tan’ or ‘cot’ term is given in the question always try to convert it into ‘sin’ and ‘cos’ terms.