A geometry problem by Mohammad Khaza

Geometry Level 2

Is the following a trigonometric identity:

1 − sin ⁡ x = ( 1 + sin ⁡ x ) ( sec ⁡ x − tan ⁡ x ) 2 1 - \sin x = ( 1 + \sin x ) ( \sec x - \tan x ) ^2

yes impossible no don't know

Mohammad Khaza by Mohammad Khaza , 20, Bangladesh

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Hana Wehbi
Jun 26, 2017

Starting with R.H.S, we have

( 1 + sin ⁡ x ) ( sec ⁡ x − tan ⁡ x ) 2 = ( 1 + sin ⁡ x ) ( 1 − sin ⁡ x cos ⁡ x ) 2 = ( 1 + sin ⁡ x ) ( 1 − sin ⁡ x ) 2 cos ⁡ 2 ( x ) = (1 + \sin x)(\sec x - \tan x)^2 = ( 1+ \sin x) (\frac{ 1 - \sin x}{ \cos x})^2= ( 1+ \sin x)\frac{(1- \sin x)^2}{\cos ^2(x)} =

( 1 + sin ⁡ x ) ( 1 − sin ⁡ x ) ( 1 − sin ⁡ x ) cos ⁡ 2 x = ( 1 − sin ⁡ 2 ( x ) ) ( 1 − sin ⁡ x ) 1 − sin ⁡ 2 ( x ) = 1 − sin ⁡ x \frac{(1+\sin x)(1-\sin x)(1-\sin x)}{\cos^2 x} = \frac{(1 - \sin^2(x))(1 -\sin x)}{1-\sin^2(x)} = 1 - \sin x which is the L.H.S.

I used these identities in my solution ( a − b ) ( a + b ) = a 2 − b 2 ( a-b)(a+b)= a^2 -b^2 and c o s 2 x + s i n 2 x = 1 cos^2 x+sin^2 x= 1

3 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...