Question

Prove that: ${(\sin A + \sec A)^2} + {(\cos A + \operatorname{cosec} A)^2} = {(1 + \sec A\cos ecA)^2}$

Answer 1

Hint : First of all we will use the expansion of the whole square formula and then simplify the equation. We will also use the equivalent trigonometric functions and replace for the resultant solution.

Complete step-by-step answer :
Take the given left hand side of the equation –
$LHS = {(\sin A + \sec A)^2} + {(\cos A + \operatorname{cosec} A)^2}$
Open the bracket using the identity for the whole square - ${(a + b)^2} = {a^2} + 2ab + {b^2}$
LHS $= {\sin ^2}A + 2\sin A\sec A + {\sec ^2}A + {\cos ^2}A + 2\cos A\operatorname{cosec} A + \cos e{c^2}A$
The above equation can be re-written as -
LHS $= \underline {{{\sin }^2}A + {{\cos }^2}A} + 2\sin A\sec A + {\sec ^2}A + 2\cos A\operatorname{cosec} A + \cos e{c^2}A$
We know that - ${\sin ^2}A + {\cos ^2}A = 1$ place in the above equation and simplify the equation –
LHS $= 1 + 2\sin A\sec A + {\sec ^2}A + 2\cos A\cos ecA + \cos e{c^2}A$
We know that the secant is the inverse of cosine function. So place, $\sec A = \dfrac{1}{{\cos A}}$ and $\cos ecA = \dfrac{1}{{\sin A}}$ in the above equation.
LHS $= 1 + 2\sin A\dfrac{1}{{\cos A}} + \dfrac{1}{{co{s^2}A}} + 2\cos A\dfrac{1}{{\sin A}} + \dfrac{1}{{{{\sin }^2}A}}$
The above equation can be arranged and re-written as –
LHS $= 1 + \dfrac{{2\sin A}}{{\cos A}} + \dfrac{1}{{co{s^2}A}} + \dfrac{{2\cos A}}{{\sin A}} + \dfrac{1}{{{{\sin }^2}A}}$
Again,
LHS $= 1 + \dfrac{1}{{{{\sin }^2}A}} + \dfrac{1}{{co{s^2}A}} + \underline {\dfrac{{2\cos A}}{{\sin A}} + \dfrac{{2\sin A}}{{\cos A}}}$
Take out the common multiple from the paired term –
LHS $= 1 + \left( {\underline {\dfrac{1}{{{{\sin }^2}A}} + \dfrac{1}{{co{s^2}A}}} } \right) + 2\left( {\underline {\dfrac{{\cos A}}{{\sin A}} + \dfrac{{\sin A}}{{\cos A}}} } \right)$
Take LCM (Least common Multiple ) in the above two paired terms and then simplify.
LHS $= 1 + \left( {\dfrac{{co{s^2}A + {{\sin }^2}A}}{{{{\sin }^2}A{{\cos }^2}A}}} \right) + 2\left( {\dfrac{{\cos A\cos A + \sin A\sin A}}{{\sin A\cos A}}} \right)$
Use the identity that - ${\sin ^2}A + {\cos ^2}A = 1$ place in the above equation and simplify the equation –
LHS $= 1 + \left( {\dfrac{1}{{{{\sin }^2}A{{\cos }^2}A}}} \right) + 2\left( {\dfrac{{co{s^2}A + {{\sin }^2}A}}{{\sin A\cos A}}} \right)$
Again using the same identity –
LHS $= 1 + \left( {\dfrac{1}{{{{\sin }^2}A{{\cos }^2}A}}} \right) + 2\left( {\dfrac{1}{{\sin A\cos A}}} \right)$
It can be re-written as –
LHS $= 1 + \left( {\dfrac{1}{{{{\sin }^2}A{{\cos }^2}A}}} \right) + \left( {\dfrac{2}{{\sin A\cos A}}} \right)$
LHS $= 1 + \left( {\dfrac{2}{{\sin A\cos A}}} \right) + \left( {\dfrac{1}{{{{\sin }^2}A{{\cos }^2}A}}} \right)$
The above equation satisfies the whole square of two terms - ${(a + b)^2} = {a^2} + 2ab + {b^2}$
LHS $= {\left( {1 + \dfrac{1}{{\sin A\cos A}}} \right)^2}$
Again convert the above equation in the form of the required answer. $\cos ecA = \dfrac{1}{{\sin A}},\sec A = \dfrac{1}{{\cos A}}$
LHS $= {\left( {1 + \cos ecA\sec A} \right)^2}$
LHS $=$ RHS
Hence, proved.

Note : Always remember that whenever we have to prove the trigonometric relation, if in between you do not find any identities and further simplifications convert all the given functions in terms of cosine and sine and then use the identity and simplify. Then for the final answer convert the solution as per requirement .