Man this one is a world of its own :D Just a quick question are you a fellow Add Math student in O levels i remember this question from back in the day :D Anyhow Lets get started

For this question we need to know the following identities:

$1+tan^{2}x=sec^2x\\\\1+cot^2x=cosec^2x\\\\sin^2x+cos^2x=1$

Lets solve the bottom most part first:

$1-\frac{1}{1-sec^2x} \\\\$

Take LCM

$1-\frac{1}{1-sec^2x} \\\\\frac{1-sec^2x-1}{1-sec^2x} \\\\\frac{-sec^2x}{1-sec^2x} \\\\\frac{-(1+tan^2x)}{-tan^2x}$

now break the LCM

$\frac{-1}{-tan^2x}+\frac{-tan^2x}{-tan^2x}\\\\\frac{1}{tan^2x}+1\\\\cot^2x+1$

because 1/tan = cot x

and furthermore,

$cot^2x+1\\cosec^2x$

now we solve the above part and replace the bottom most part that we solved with $cosec^2x$

$\frac{1}{1-\frac{1}{cosec^2x} } \\\\\frac{1}{1-sin^2x} \\\\\frac{1}{cos^2x}\\\\sec^2x$

Hence proved! :D

4 0

3 years ago

Persons younger than 18 years old are not eligible to vote. Let a be age in years.

BaLLatris [955]

D because a has to be younger than 18 and it has to be positive because we are never a negative age. do you understand when i mean?

3 0

3 years ago

Read 2 more answers

Complete the table for the given rule. Rule: y = x/2 - 1 X Y 4 = 6 = 8 =

Veseljchak [2.6K]

$\bf \begin{array}{|cc|ll} \cline{1-2} x&y\\ \cline{1-2} 4&\stackrel{\frac{4}{2}-1}{1}\\&\\ 6&\stackrel{\frac{6}{2}-1}{2}\\&\\ 8&\stackrel{\frac{8}{2}-1}{3}\\ &\\ \cline{1-2} \end{array}$

5 0

3 years ago

Find the greatest common factor of 18, 36,18,

Vikki [24]

Answer:

Step-by-step explanation:

18=2(3)3

36=2(2)3(3)

The GCF will be the product of their common factors, specifically 2(3)3=18

5 0

2 years ago