Question

Find the value of k for the given equation: \[k+1={{\sec }^{2}}\theta \left( 1+\sin \theta \right)\left( 1-\sin \theta \right)\] .

Answer 1

Hint: First of all, consider the given expression and convert \[\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)\text{ into }1-{{\sin }^{2}}\theta \] by using \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\]. Now, use \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\text{ and }\sec x=\dfrac{1}{\cos x}\] to simplify the equation and get the value of k.
Complete step-by-step answer:
We are given that \[k+1={{\sec }^{2}}\theta \left( 1+\sin \theta \right)\left( 1-\sin \theta \right)\], we have to find the value of k. Let us consider the expression given in the question.
\[k+1={{\sec }^{2}}\theta \left( 1+\sin \theta \right)\left( 1-\sin \theta \right)\]
We know that \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\]. By using this in the above expression, we get,
\[\left( k+1 \right)={{\sec }^{2}}\theta \left[ {{\left( 1 \right)}^{2}}-{{\left( \sin \theta \right)}^{2}} \right]\]
We can also write the above equation as
\[\left( k+1 \right)={{\sec }^{2}}\theta \left( 1-{{\sin }^{2}}\theta \right)\]
We know that
\[{{\sin }^{2}}x+{{\cos }^{2}}x=1\text{ or }1-{{\sin }^{2}}x={{\cos }^{2}}x\]
By using this in the above equation, we get,
\[\left( k+1 \right)=\left( {{\sec }^{2}}\theta \right)\left( {{\cos }^{2}}\theta \right)\]
We know that \[\sec x=\dfrac{1}{\cos x}\]. By using this in the above equation, we get,
\[\left( k+1 \right)={{\left( \dfrac{1}{\cos \theta } \right)}^{2}}\left( {{\cos }^{2}}\theta \right)\]
\[\left( k+1 \right)=\left( \dfrac{1}{\cos \theta } \right).{{\cos }^{2}}\theta \]
By canceling the like terms from the above equation, we get,
k + 1 = 1
By subtracting 1 from both the sides of the above equation, we get,
k + 1 – 1 = 1 – 1
k = 0
So, we get the value of k as 0.

Note: It is always advisable to convert and simplify the equation in terms of \[\sin \theta \text{ and }\cos \theta \]. Students can also cross-check their answer by substituting k = 1 in the given equation and checking if LHS = RHS or not. In this question, many students make this mistake of writing \[\left( 1-\sin \theta \right)\left( 1+\sin \theta \right)\text{ as }{{\sin }^{2}}\theta -1\] which is wrong. So, this must be taken care of by carefully taking a and b in the formula \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\].