Question

${{\left( \cos A+\sin A \right)}^{2}}-{{\left( \cos A-\sin A \right)}^{2}}$ is equal to
[a] -1
[b] 2
[c] 0
[d] None of the above.

Answer 1

Hint: Use the identity ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ to simplify the expression. Use $\sin 2A=2\sin A\cos A$ and hence find the simplified expression. Compare with the given options and check which of the options is correct.

Complete step-by-step answer:
We have
${{\left( \cos A+\sin A \right)}^{2}}-{{\left( \cos A-\sin A \right)}^{2}}$
We know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
Taking a = cosA+sinA and b = cosA-sinA in the above expression, we get
${{\left( \cos A+\sin A \right)}^{2}}-{{\left( \cos A-\sin A \right)}^{2}}=\left( \cos A+\sin A+\cos A-\sin A \right)\left( \cos A+\sin A-\left( \cos A-\sin A \right) \right)$
Simplifying the above expression, we get
${{\left( \cos A+\sin A \right)}^{2}}-{{\left( \cos A-\sin A \right)}^{2}}=\left( 2\cos A \right)\left( 2\sin A \right)=2\times 2\sin A\cos A$
We know that $\sin 2x=2\sin x\cos x$
Hence we have
${{\left( \cos A+\sin A \right)}^{2}}-{{\left( \cos A-\sin A \right)}^{2}}=2\sin 2A$
Now since 2sin2A is not independent of A, none of the given options is correct.
Hence option [d] is correct.

Note: Alternatively, we have
${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Put a = cosA and b = sinA, we get
${{\left( \cos A+\sin A \right)}^{2}}={{\cos }^{2}}A+2\sin A\cos A+{{\sin }^{2}}A$
We know that $2\sin A\cos A=\sin 2A$
Hence we have
${{\left( \cos A+\sin A \right)}^{2}}={{\cos }^{2}}A+{{\sin }^{2}}A+\sin 2A$
We know that ${{\cos }^{2}}x+{{\sin }^{2}}x=1$
Hence we have
${{\left( \cos A+\sin A \right)}^{2}}=1+\sin 2A\text{ (i)}$
We know that
${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
Put a = cosA and b = sinA, we get
${{\left( \cos A-\sin A \right)}^{2}}={{\cos }^{2}}A-2\sin A\cos A+{{\sin }^{2}}A$
We know that $2\sin A\cos A=\sin 2A$
Hence we have
${{\left( \cos A-\sin A \right)}^{2}}={{\cos }^{2}}A+{{\sin }^{2}}A-\sin 2A$
We know that ${{\cos }^{2}}x+{{\sin }^{2}}x=1$
Hence we have
${{\left( \cos A-\sin A \right)}^{2}}=1-\sin 2A\text{ (ii)}$
Subtracting equation (ii) from equation (i), we get
${{\left( \cos A+\sin A \right)}^{2}}-{{\left( \cos A-\sin A \right)}^{2}}=\left( 1+\sin 2A \right)-\left( 1-\sin 2A \right)$
Expanding the expression on the RHS, we get
${{\left( \cos A+\sin A \right)}^{2}}-{{\left( \cos A-\sin A \right)}^{2}}=1+\sin 2A-1+\sin 2A=2\sin 2A$, which is the same as obtained above.
Hence option [d] is correct.
[ii] Alternatively, we know that ${{\left( a+b \right)}^{2}}-{{\left( a-b \right)}^{2}}=4ab$
Put a = cosA and b = cosA, we have
${{\left( \cos A+\sin A \right)}^{2}}-{{\left( \cos A-\sin A \right)}^{2}}=4\sin A\cos A$
Using $\sin A\cos A=\dfrac{\sin 2A}{2}$, we get
${{\left( \cos A+\sin A \right)}^{2}}-{{\left( \cos A-\sin A \right)}^{2}}=4\times \dfrac{\sin 2A}{2}=2\sin 2A$, which is the same as obtained above.
Hence option [d] is correct.