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Question: Prove: \[{\left( {\sin A + \cos ecA} \right)^2} + {\left( {\cos A + \sec A} \right)^2} = 7 + {\tan ^...

Prove: (sinA+cosecA)2+(cosA+secA)2=7+tan2A+cot2A{\left( {\sin A + \cos ecA} \right)^2} + {\left( {\cos A + \sec A} \right)^2} = 7 + {\tan ^2}A + {\cot ^2}A ?

Explanation

Solution

The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as cos2x+sin2x=1{\cos ^2}x + {\sin ^2}x = 1 and tan2x+1=sec2x{\tan ^2}x + 1 = {\sec ^2}x. Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem and proving the result given to us.

Complete step by step solution:
In the given problem, we have to prove a trigonometric equality that can be further used in many questions and problems as a direct result and has wide ranging applications. For proving the desired result, we need to have a good grip over the basic trigonometric formulae and identities.
Now, we need to make the left and right sides of the equation equal.
R.H.S. =(sinA+cosecA)2+(cosA+secA)2 = {\left( {\sin A + \cos ecA} \right)^2} + {\left( {\cos A + \sec A} \right)^2}
Now, we have to first simplify the expression by opening the brackets and evaluating the whole squares. So, we get,
sin2A+cosec2A+2sinAcosecA+cos2A+sec2A+2cosAsecA\Rightarrow {\sin ^2}A + \cos e{c^2}A + 2\sin A\cos ecA + {\cos ^2}A + {\sec ^2}A + 2\cos A\sec A
Now, we know that secx and cosx are reciprocals of each other. Hence, we have secx=1cosx\sec x = \dfrac{1}{{\cos x}}. So, we get, cosxsecx=1\cos x\sec x = 1. Similarly, we get cosecxsinx=1\cos ecx\sin x = 1. So, substituting these values, we get,
sin2A+cosec2A+2(1)+cos2A+sec2A+2(1)\Rightarrow {\sin ^2}A + \cos e{c^2}A + 2\left( 1 \right) + {\cos ^2}A + {\sec ^2}A + 2\left( 1 \right)
4+(sin2A+cos2A)+cosec2A+sec2A\Rightarrow 4 + \left( {{{\sin }^2}A + {{\cos }^2}A} \right) + \cos e{c^2}A + {\sec ^2}A
Now, we have to use trigonometric identities such as cos2x+sin2x=1{\cos ^2}x + {\sin ^2}x = 1, tan2x+1=sec2x{\tan ^2}x + 1 = {\sec ^2}x and cot2x+1=cosec2x{\cot ^2}x + 1 = co{\sec ^2}x.
Now, we get,
4+1+(cot2A+1)+(tan2A+1)\Rightarrow 4 + 1 + \left( {{{\cot }^2}A + 1} \right) + \left( {{{\tan }^2}A + 1} \right)
Now, simplifying the expression, we get,
7+cot2A+tan2A\Rightarrow 7 + {\cot ^2}A + {\tan ^2}A
Now, L.H.S=7+cot2A+tan2A = 7 + {\cot ^2}A + {\tan ^2}A
As the left side of the equation is equal to the right side of the equation, we have,
(sinA+cosecA)2+(cosA+secA)2=7+tan2A+cot2A{\left( {\sin A + \cos ecA} \right)^2} + {\left( {\cos A + \sec A} \right)^2} = 7 + {\tan ^2}A + {\cot ^2}A
Hence, Proved.

Note: Given problem deals with Trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such type of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations.