First slide

Trigonometric Identities

Question

Let us consider the equation $\frac{\cos^{4} x}{a} + \frac{\sin^{4} x}{b} = \frac{1}{a + b}, x \in [0, \frac{π}{2}], a, b > 0$

Moderate

Question

For the given equation, which of the following is correct

A

$\frac{\sin^{4} x}{b} = \frac{\cos^{4} x}{a}$
B

$\frac{\sin x}{a} = \frac{\cos x}{b}$
C

$\frac{\sin^{4} x}{b^{2}} = \frac{\cos^{4} x}{a^{2}}$
D

$\frac{\sin^{2} x}{a} = \frac{\cos^{2} x}{b}$

Solution

$\begin{array}{l} We have, \frac{\cos^{4} x}{a} + \frac{\sin^{4} x}{b} = \frac{1}{a + b} = \frac{\cos^{2} x + \sin^{2} x}{a + b} \\ \Rightarrow \cos^{2} x (\frac{\cos^{2} x}{a} - \frac{1}{a + b}) = \sin^{2} x (\frac{1}{a + b} - \frac{\sin^{2} x}{b}) \\ \Rightarrow \frac{\cos^{2} x (\frac{{bcos}^{2} x - {asin}^{2} x}{a (a + b)})}{a} = \sin^{2} x (\frac{{bcos}^{2} x - {asin}^{2} x}{b (a + b)}) \\ \Rightarrow \frac{1 - \sin^{2} x}{a} = \frac{\sin^{2} x}{b} \\ \Rightarrow S i n^{2} x = \frac{b}{a + b} and \cos^{2} x = \frac{a}{a + b} \end{array}$

Question

The value of sin²x in terms of a and b is

A

$\sqrt{ab}$
B

$\frac{b}{a + b}$
C

$\frac{b^{2} - a^{2}}{a^{2} + b^{2}}$
D

$\frac{a^{2} + b^{2}}{b^{2} - a^{2}}$

Solution

$\begin{array}{l} We have, \frac{\cos^{4} x}{a} + \frac{\sin^{4} x}{b} = \frac{1}{a + b} = \frac{\cos^{2} x + \sin^{2} x}{a + b} \\ \Rightarrow \cos^{2} x (\frac{\cos^{2} x}{a} - \frac{1}{a + b}) = \sin^{2} x (\frac{1}{a + b} - \frac{\sin^{2} x}{b}) \\ \Rightarrow \frac{\cos^{2} x (\frac{{bcos}^{2} x - {asin}^{2} x}{a (a + b)})}{a} = \sin^{2} x (\frac{{bcos}^{2} x - {asin}^{2} x}{b (a + b)}) \\ \Rightarrow \frac{1 - \sin^{2} x}{a} = \frac{\sin^{2} x}{b}) \\ \Rightarrow \sin^{2} x = \frac{b}{a + b} and \cos^{2} x = \frac{a}{a + b} \end{array}$

Question

The value of $\frac{\sin^{8} x}{b^{3}} + \frac{\cos^{8} x}{a^{3}}$ is

A

$\frac{1}{(a + b)^{2}}$
B

$\frac{1}{(a + b)^{3}}$
C

$\frac{1}{(a + b)^{4}}$
D

$\frac{1}{a^{3} + b^{3}}$

Solution

$\begin{array}{l} We have, \frac{\cos^{4} x}{a} + \frac{\sin^{4} x}{b} = \frac{1}{a + b} = \frac{\cos^{2} x + \sin^{2} x}{a + b} \\ \Rightarrow \cos^{2} x (\frac{\cos^{2} x}{a} - \frac{1}{a + b}) = \sin^{2} x (\frac{1}{a + b} - \frac{\sin^{2} x}{b}) \\ \Rightarrow \frac{\cos^{2} x (\frac{{bcos}^{2} x - {asin}^{2} x}{a (a + b)})}{a} = \sin^{2} x (\frac{{bcos}^{2} x - {asin}^{2} x}{b (a + b)}) \\ \Rightarrow \frac{1 - \sin^{2} x}{a} = \frac{\sin^{2} x}{b}) \\ \Rightarrow \frac{\sin^{2} x}{b} = \frac{b}{a + b} and \cos^{2} x = \frac{a}{a + b} \\ \Rightarrow b^{3} + \frac{\sin^{2} x^{3} x}{a^{3}} = \frac{b^{4}}{b^{3} (a + b)^{4}} + \frac{a^{3}}{a^{3} (a + b)^{4}} \end{array}$

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