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

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

Explanation

Solution

Hint: First of all, consider the given expression and convert (1+sin⁡θ)(1−sin⁡θ) into 1−sin⁡2θ\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)\text{ into }1-{{\sin }^{2}}\theta by using (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}. Now, use sin⁡2x+cos⁡2x=1 and sec⁡x=1cos⁡x{{\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⁡2θ(1+sin⁡θ)(1−sin⁡θ)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⁡2θ(1+sin⁡θ)(1−sin⁡θ)k+1={{\sec }^{2}}\theta \left( 1+\sin \theta \right)\left( 1-\sin \theta \right)
We know that (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}. By using this in the above expression, we get,
(k+1)=sec⁡2θ[(1)2−(sin⁡θ)2]\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
(k+1)=sec⁡2θ(1−sin⁡2θ)\left( k+1 \right)={{\sec }^{2}}\theta \left( 1-{{\sin }^{2}}\theta \right)
We know that
sin⁡2x+cos⁡2x=1 or 1−sin⁡2x=cos⁡2x{{\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,
(k+1)=(sec⁡2θ)(cos⁡2θ)\left( k+1 \right)=\left( {{\sec }^{2}}\theta \right)\left( {{\cos }^{2}}\theta \right)
We know that sec⁡x=1cos⁡x\sec x=\dfrac{1}{\cos x}. By using this in the above equation, we get,
(k+1)=(1cos⁡θ)2(cos⁡2θ)\left( k+1 \right)={{\left( \dfrac{1}{\cos \theta } \right)}^{2}}\left( {{\cos }^{2}}\theta \right)
(k+1)=(1cos⁡θ).cos⁡2θ\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⁡θ and cos⁡θ\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 (1−sin⁡θ)(1+sin⁡θ) as sin⁡2θ−1\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 (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}.