Question

Solve the following ${{\sin }^{2}}x+{{\cos }^{2}}y=2{{\sec }^{2}}z$ for x, y and z.
A. $x=\left( 2m+1 \right)\dfrac{\pi }{2}$, where m is an integer.
B. $y=n\pi$, where n is an integer.
C. $z=t\pi$, where t is an integer.
D. $x=\left( 2m+1 \right)\dfrac{\pi }{4}$, where m is an integer.

Answer 1

Hint: To solve this question, we should have some knowledge of some of the general solutions of sin and cos functions. Like, if $\sin \theta =0$, then $\theta =n\pi$, if $\cos \theta =0$, then $\theta =m\pi +\dfrac{\pi }{2}$ and if, $\sin \theta =\alpha$ then $\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha$. We should also know that $-1\le \cos \theta \le 1$ and $-1\le \sin \theta \le 1$ and $\sec \theta \ge 1$ or $sec\theta \le -1$. By using these identities, we can solve this question.

Complete step-by-step answer:
In this question, we have been asked to solve the equation, ${{\sin }^{2}}x+{{\cos }^{2}}y=2{{\sec }^{2}}z$ for x, y and z. To solve this question, we should know that, $-1\le \sin \theta \le 1$ or we can say $0\le {{\sin }^{2}}x\le 1$ and that, $-1\le \cos \theta \le 1$ or we can say $0\le {{\cos }^{2}}y\le 1$. So, we can say that the minimum possible value of ${{\sin }^{2}}x+{{\cos }^{2}}y$ is 0 and the maximum possible value is 2. So, we can say,
$0\le {{\sin }^{2}}x+{{\cos }^{2}}y\le 2\ldots \ldots \ldots \left( i \right)$
We also know that $\left| \sec \theta \right|\ge 1$. So, we can say that ${{\sec }^{2}}\theta \ge 1$. So, we can write,
$2{{\sec }^{2}}\theta \ge 2\ldots \ldots \ldots \left( ii \right)$
From equation (i) and (ii), we can say that the equality ${{\sin }^{2}}x+{{\cos }^{2}}y=2{{\sec }^{2}}z$ exists only when ${{\sin }^{2}}x+{{\cos }^{2}}y=2$ and $2{{\sec }^{2}}z=2$. So, to satisfy this condition, we can say, ${{\sin }^{2}}x=1,{{\cos }^{2}}y=1,{{\sec }^{2}}z=1$. And we know that $\sec \theta =\dfrac{1}{\cos \theta }$. So, for $\sec \theta =1$, we will get $\cos \theta =1$. Hence, we can say that, ${{\sin }^{2}}x=1,{{\cos }^{2}}y=1,{{\cos }^{2}}z=1$.
Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. So, if we put ${{\sin }^{2}}\theta =1$, then we get $\cos \theta =0$ and if we put ${{\cos }^{2}}\theta =1$, then we get $\sin \theta =0$. So, we can write the above equalities as, $\cos x=0,\sin y=0,\sin z=0$.
Now, we know that if $\cos \theta =0$, then $\theta =m\pi +\dfrac{\pi }{2}$. So, we can say that, $x=2m\pi +\dfrac{\pi }{2}\Rightarrow x=\left( 2m+1 \right)\dfrac{\pi }{2}$, where m is an integer.
We also know that if $\sin \theta =0$, then $\theta =n\pi$. So, we can say that, $y=n\pi$ and $z=t\pi$, where n and t are integers.
Hence, we can say that $x=\left( 2m+1 \right)\dfrac{\pi }{2},m\in z;y=n\pi ,n\in z$ and $z=t\pi ,t\in z$ are the solutions for ${{\sin }^{2}}x+{{\cos }^{2}}y=2{{\sec }^{2}}z$.
Therefore, options A, B and C are the correct answers.

Note: While solving this question, one can make a mistake by writing ${{\sin }^{2}}x+{{\cos }^{2}}y=1$,which would be totally incorrect as this property satisfies for same angles, that is, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. One should also remember that $\cos \theta =\dfrac{1}{\sec \theta }$. This question has multiple correct answers, so one must check carefully not to miss out any option.