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8. Introduction to Trigonometry
easy
$\frac{\cos ^{2} 40+\cos ^{2} 50}{\sin ^{2} 40+\sin ^{2} 50}=\ldots \ldots \ldots \ldots$
A
$2$
B
$4$
C
$1$
D
$0$
Solution
$\cos ^{2} 50=\sin ^{2}(90-50)=\sin ^{2} 40$
Similarly, $\sin ^{2} 50=\cos ^{2} 40$
Now, $\frac{\cos ^{2} 40+\cos ^{2} 50}{\sin ^{2} 40+\sin ^{2} 50}=\frac{\cos ^{2} 40+\sin ^{2} 40}{\sin ^{2} 40+\cos ^{2} 40}=\frac{1}{1}=1$
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
