MCQ
$\cot (45^\circ + \theta )\cot (45^\circ - \theta ) = $
- A$-1$
- B$0$
- ✓$1$
- D$\infty $
$= \tan (90^\circ - 45^\circ - \theta )\cot (45^\circ - \theta )$
$ = \tan (45^\circ - \theta )\cot (45^\circ - \theta ) = 1$.
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$y=\log _{10} x+\log _{10} x^{1 / 3}+\log _{10} x^{1 / 9}+\ldots . .$ upto $\infty$ terms and $\frac{2+4+6+\ldots+2 \mathrm{y}}{3+6+9+\ldots+3 \mathrm{y}}=\frac{4}{\log _{10} \mathrm{x}}$, then the ordered pair $(x, y)$ is equal to :