Question

If $\sin \theta + {\sin ^2}\theta = 1$ and $a{\cos ^{12}}\theta + b{\cos ^{10}}\theta + c{\cos ^8}\theta + d{\cos ^6}\theta - 1 = 0$, then $\dfrac{{b + c}}{{a + d}} =$?
A) $1$
B) $2$
C) $- 2$
D) $3$

Answer 1

In the given question, we are given that $\sin \theta + {\sin ^2}\theta = 1$, $a{\cos ^{12}}\theta + b{\cos ^{10}}\theta + c{\cos ^8}\theta + d{\cos ^6}\theta - 1 = 0$ and we need to find the value of$\dfrac{{b + c}}{{a + d}}$ . At first, we will try to convert both the equations in terms of one trigonometric function only, here we will try to convert the second equation in terms of $\sin e$ function. After this, we will use algebraic identity and compare the LHS and RHS of the equation to get the values of $a,b,c,d$ and substitute them in $\dfrac{{b + c}}{{a + d}}$ to get our answer.

Complete step by step answer:
Given, $\sin \theta + {\sin ^2}\theta = 1$
$\Rightarrow \sin \theta = 1 - {\sin ^2}\theta$
As we know ${\sin ^2}x + {\cos ^2}x = 1$. Therefore, we get
$\Rightarrow \sin \theta = {\cos ^2}\theta ...\left( i \right)$
We have, $a{\cos ^{12}}\theta + b{\cos ^{10}}\theta + c{\cos ^8}\theta + d{\cos ^6}\theta - 1 = 0$
This can also be written as:
$\Rightarrow a{\left( {{{\cos }^2}\theta } \right)^6} + b{\left( {{{\cos }^2}\theta } \right)^5} + c{\left( {{{\cos }^2}\theta } \right)^4} + d{\left( {{{\cos }^2}\theta } \right)^3} - 1 = 0$
As we know $\sin \theta = {\cos ^2}\theta$ from $\left( i \right)$. Therefore, we get
$\Rightarrow a{\left( {\sin \theta } \right)^6} + b{\left( {\sin \theta } \right)^5} + c{\left( {\sin \theta } \right)^4} + d{\left( {\sin \theta } \right)^3} - 1 = 0$
This can also be written as,
$\Rightarrow a{\sin ^6}\theta + b{\sin ^5}\theta + c{\sin ^4}\theta + d{\sin ^3}\theta = 1...\left( {ii} \right)$
As we have, $\sin \theta + {\sin ^2}\theta = 1$
On cubing both sides, we get
$\Rightarrow {\left( {\sin \theta + {{\sin }^2}\theta } \right)^3} = {1^3}$
$\Rightarrow {\left( {\sin \theta + {{\sin }^2}\theta } \right)^3} = 1$
On replacing $1$ by ${\left( {\sin \theta + {{\sin }^2}\theta } \right)^3}$ in equation $\left( {ii} \right)$, we get
$\Rightarrow a{\sin ^6}\theta + b{\sin ^5}\theta + c{\sin ^4}\theta + d{\sin ^3}\theta = {\left( {\sin \theta + {{\sin }^2}\theta } \right)^3}$
Now, we will solve the RHS of the above written equation using ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$. Therefore, we get
$\Rightarrow a{\sin ^6}\theta + b{\sin ^5}\theta + c{\sin ^4}\theta + d{\sin ^3}\theta = {\sin ^3}\theta + {\left( {{{\sin }^2}\theta } \right)^3} + 3\sin \theta \times {\sin ^2}\theta \left( {\sin \theta + {{\sin }^2}\theta } \right)$
On multiplication of terms written in LHS, we get
$\Rightarrow a{\sin ^6}\theta + b{\sin ^5}\theta + c{\sin ^4}\theta + d{\sin ^3}\theta = {\sin ^3}\theta + {\sin ^6}\theta + 3{\sin ^4}\theta + 3{\sin ^5}\theta$
On comparing the coefficients of ${\sin ^6}\theta ,{\sin ^5}\theta ,{\sin ^4}\theta ,{\sin ^3}\theta$, we get
$a = 1$, $b = 3$, $c = 3$ and $d = 1$.
Now, we will substitute these values in $\dfrac{{b + c}}{{a + d}}$.
$\Rightarrow \dfrac{{b + c}}{{a + d}} = \dfrac{{3 + 3}}{{1 + 1}}$
On addition of terms, we get
$\Rightarrow \dfrac{{b + c}}{{a + d}} = \dfrac{6}{2}$
$\therefore \dfrac{{b + c}}{{a + d}} = 3$
Thus, the value $\dfrac{{b + c}}{{a + d}}$ is $3$.
Therefore, option (D) is correct.

Note:
To solve this type of question, try to convert both the equations in the same function only. One must know trigonometric formulae and algebraic identities to solve this type of question. Here, we have used product rule to multiply the $\sin e$ functions, according to product rule, ${a^m} \times {a^n} = {a^{m + n}}$, the product of multiplication of exponents with the same base is equal to the sum of their powers with same base. In order to solve questions of this type the key is to identify the nature of the given question and use the formula that fits and use it accordingly. We should take care of the calculations so as to be sure of our final answer.