MCQ
If $\tan\theta=\frac{3}{4},$ then $\cos^2\theta-\sin^2\theta=$
- ✓$\frac{7}{25}$
- B$1$
- C$\frac{-7}{25}$
- D$\frac{4}{25}$
✓
Answer
Correct option: A.
$\frac{7}{25}$
We have,
$\tan\theta= \frac{3}{4}$
In $\triangle \text{ABC},$
$\text{AC}^2=\text {AB}^2+\text{BC}^2$
$\Rightarrow \text {AC}^2=(3)^2+(4)^2$
$\Rightarrow \text {AC}^2=9+16$
$\Rightarrow \text {AC}^2=25$
$\Rightarrow \text {AC}=5$
$\therefore \sin \theta=\frac{3}{5}$ and $\cos\theta=\frac{4}{5}$
Now, $\cos^2\theta-\sin^2 =\Big(\frac{4}{5}\Big)^2-\Big(\frac{3}{5}\Big)^2$
$=\frac{16}{25}- \frac{9}{25}$
$=\frac{16-9}{25}$
$=\frac{7}{25}$
Hence the correct option is $(a)$
$\tan\theta= \frac{3}{4}$
In $\triangle \text{ABC},$
$\text{AC}^2=\text {AB}^2+\text{BC}^2$
$\Rightarrow \text {AC}^2=(3)^2+(4)^2$
$\Rightarrow \text {AC}^2=9+16$
$\Rightarrow \text {AC}^2=25$
$\Rightarrow \text {AC}=5$
$\therefore \sin \theta=\frac{3}{5}$ and $\cos\theta=\frac{4}{5}$
Now, $\cos^2\theta-\sin^2 =\Big(\frac{4}{5}\Big)^2-\Big(\frac{3}{5}\Big)^2$
$=\frac{16}{25}- \frac{9}{25}$
$=\frac{16-9}{25}$
$=\frac{7}{25}$
Hence the correct option is $(a)$
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