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8. Introduction to Trigonometry
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Prove that $\sin ^{6} \theta+\cos ^{6} \theta+3 \sin ^{2} \theta \cos ^{2} \theta=1$
Option A
Option B
Option C
Option D
Solution
We know that $\sin ^{2} \theta+\cos ^{2} \theta=1$
Therefore, $\quad\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{3}=1$
or, $\left(\sin ^{2} \theta\right)^{3}+\left(\cos ^{2} \theta\right)^{3}+3 \sin ^{2} \theta \cos ^{2} \theta\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=1$
or, $\sin ^{6} \theta+\cos ^{6} \theta+3 \sin ^{2} \theta \cos ^{2} \theta=1$
Standard 10
Mathematics
Similar Questions
Which of the following pair is correct for trigonometric inter-relationship ?
| $1 .$ $\cos \theta$ | $a.$ $\frac{\cos \theta}{\sin \theta}$ |
| $2.$ $\tan \theta$ | $b.$ $\frac{1}{\operatorname{coses} \theta}$ |
| $3 .$ $\cot \theta$ | $c.$ $\frac{1}{\sec \theta}$ |
| $4.$ $\sin \theta$ | $d.$ $\frac{1}{\cot \theta}$ |
| $e.$ $\sin \theta \cdot \cos \theta$ |
easy
