Question

Given that a=sec x + cosec x and b=sec x – cosec x, show that \[{{a}^{2}}+{{b}^{2}}=2{{\sec }^{2}}x\cdot \cos e{{c}^{2}}x\]?

Answer 1

Hint: This type of problem is based on the concept of trigonometry. We have been given the values of a and b. square a and b using the identities and formulas. Add \[{{a}^{2}}\] and \[{{b}^{2}}\]. Do necessary calculations and group the common terms. We know that \[\cos ecx=\dfrac{1}{\sin x}\] and \[\sec x=\dfrac{1}{\cos x}\]. Substitute these values in the expression and simplify the expression with the trigonometric identity\[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]. Do necessary calculation and find the value of \[{{a}^{2}}+{{b}^{2}}\].

Complete step by step solution:
According to the question, we are asked to find the value of \[{{a}^{2}}+{{b}^{2}}\].
We have been given the values of a=sec x + cosec x ------------(1)
And b=sec x – cosec x. ----------(2)
Let us first consider equation (1).
Let us square on both the sides of the equation (1).
\[\Rightarrow {{a}^{2}}={{\left( \sec x+\cos ecx \right)}^{2}}~~\]
We know that \[{{\left( A+B \right)}^{2}}={{A}^{2}}+2AB+{{B}^{2}}\]. Here, A=sec x and B=cosec x.
On substituting the values, we get
\[{{a}^{2}}={{\left( \sec x \right)}^{2}}+2\sec x\cdot \cos ecx~+{{\left( \cos ecx \right)}^{2}}\]
On further simplification, we get
\[{{a}^{2}}={{\sec }^{2}}x+2\sec x\cdot \cos ecx~+\cos e{{c}^{2}}x\] ------------(3)
Let us now consider equation (2).
Let us square on both the sides of the equation (2).
\[\Rightarrow {{b}^{2}}={{\left( \sec x-\cos ecx \right)}^{2}}~~\]
We know that \[{{\left( A-B \right)}^{2}}={{A}^{2}}-2AB+{{B}^{2}}\]. Here, A=sec x and B=cosec x.
On substituting the values, we get
\[{{b}^{2}}={{\left( \sec x \right)}^{2}}-2\sec x\cdot \cos ecx~+{{\left( \cos ecx \right)}^{2}}\]
On further simplification, we get
\[{{b}^{2}}={{\sec }^{2}}x-2\sec x\cdot \cos ecx~+\cos e{{c}^{2}}x\] ------------(4)
Let us now add equations (3) and (4).
\[\Rightarrow {{a}^{2}}+{{b}^{2}}={{\sec }^{2}}x+2\sec x\cdot \cos ecx~+\cos e{{c}^{2}}x+{{\sec }^{2}}x-2\sec x\cdot \cos ecx~+\cos e{{c}^{2}}x\]
Let us group all the similar terms.
\[{{a}^{2}}+{{b}^{2}}={{\sec }^{2}}x+{{\sec }^{2}}x+2\sec x\cdot \cos ecx~-2\sec x\cdot \cos ecx~+\cos e{{c}^{2}}x+\cos e{{c}^{2}}x\]
We know that terms with same magnitude and opposite signs cancel out. On cancelling \[2\sec x\cdot \cos ecx~\], we get
\[{{a}^{2}}+{{b}^{2}}={{\sec }^{2}}x+{{\sec }^{2}}x+\cos e{{c}^{2}}x+\cos e{{c}^{2}}x\]
On further simplification, we get
\[{{a}^{2}}+{{b}^{2}}=2{{\sec }^{2}}x+2\cos e{{c}^{2}}x\]
Let us take 2 common out of the two terms from RHS.
\[\Rightarrow {{a}^{2}}+{{b}^{2}}=2\left( {{\sec }^{2}}x+\cos e{{c}^{2}}x \right)\]
We know that \[\cos ecx=\dfrac{1}{\sin x}\] and \[\sec x=\dfrac{1}{\cos x}\].
On substituting in the above expression, we get
\[\Rightarrow {{a}^{2}}+{{b}^{2}}=2\left( {{\left( \dfrac{1}{\sin x} \right)}^{2}}+{{\left( \dfrac{1}{\cos x} \right)}^{2}} \right)\]
On further simplification, we get
\[{{a}^{2}}+{{b}^{2}}=2\left( \dfrac{1}{{{\sin }^{2}}x}+\dfrac{1}{{{\cos }^{2}}x} \right)\]
Let us take LCM in the denominator.
\[\Rightarrow {{a}^{2}}+{{b}^{2}}=2\left( \dfrac{{{\cos }^{2}}x+{{\sin }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x} \right)\]
We can write the expression as
\[{{a}^{2}}+{{b}^{2}}=2\left( \dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x} \right)\]
We know that \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]. Using this identity, we get
\[{{a}^{2}}+{{b}^{2}}=2\left( \dfrac{1}{{{\sin }^{2}}x{{\cos }^{2}}x} \right)\]
We can write the expression as
\[{{a}^{2}}+{{b}^{2}}=2\left( \dfrac{1}{{{\sin }^{2}}x}\cdot \dfrac{1}{{{\cos }^{2}}x} \right)\]
\[\Rightarrow {{a}^{2}}+{{b}^{2}}=2\left( {{\left( \dfrac{1}{\sin x} \right)}^{2}}\cdot {{\left( \dfrac{1}{\cos x} \right)}^{2}} \right)\]
But, we know that \[\cos ecx=\dfrac{1}{\sin x}\] and \[\sec x=\dfrac{1}{\cos x}\].
On substituting, we get
\[{{a}^{2}}+{{b}^{2}}=2\left( {{\left( \cos ecx \right)}^{2}}\cdot {{\left( \sec x \right)}^{2}} \right)\]
\[\Rightarrow {{a}^{2}}+{{b}^{2}}=2\left( {{\sec }^{2}}x\cdot \cos e{{c}^{2}}x \right)\]
\[\therefore {{a}^{2}}+{{b}^{2}}=2{{\sec }^{2}}x\cdot \cos e{{c}^{2}}x\]
Hence, we have proved that \[{{a}^{2}}+{{b}^{2}}=2{{\sec }^{2}}x\cdot \cos e{{c}^{2}}x\].

Note: We should know the trigonometric identities to solve this type of problems. Be careful with the conversion. We have to check whether the value given in the question matches with the value obtained. If not, the answer is incorrect. Avoid calculation mistakes based on sign conventions.