Question 66

If $$\sin x = \frac{4}{5}$$, then $$\frac{Tan x}{Cot x}$$ = ?

Solution

$$\frac{Tan x}{Cot x}=Tan^2 x$$ {$$Cot x=\frac{1}{Tan x}$$}

We know $$Tan x=\frac{Sin x}{cos x}$$

$$=\frac{Sin^2x}{Cos^2x}$$-------1

Given $$Sinx=\frac{4}{5}$$

cos x = base/hypotenuse

$$base^2=hypotenuse^2-Perpendicular^2$$

=$$5^2-4^2=25-16=9$$

base=3, Cos x=$$\frac{3}{5}$$

Put in equation one

$$=\frac{(\frac{4}{5})^2}{(\frac{3}{5})^2}$$

$$=\frac{16}{9}$$

Create a FREE account and get:

Download RRB Study Material PDF
45+ RRB previous papers with solutions PDF
300+ Online RRB Tests for Free

CAT Quant Questions | CAT Quantitative Ability

CAT Venn Diagrams Questions CAT Functions, Graphs and Statistics Questions CAT Coordinate Geometry Questions CAT Profit And Loss Questions CAT Quadratic Equations Questions

CAT DILR Questions | LRDI Questions For CAT

CAT Puzzles Questions CAT Arrangement Questions CAT Venn Diagrams Questions CAT Data Interpretation Basics Questions CAT Miscellaneous LR Questions

CAT Verbal Ability Questions | VARC Questions For CAT

CAT Verbal Ability Questions CAT Reading Comprehension Questions CAT Critical Reasoning Questions CAT Grammar and Sentence Correction Questions CAT Miscellaneous Questions

Follow us on

Boost your Prep!

Download App