Question

Verify this trigonometric equation...- tan^2x + sec^2x = 1?

Answer 1

- tan^2x + sec^2x=1
or sec^2x- tan^2x =1
consider the following triangle ABC in which angle B=90 degrees
and angle C=x

Answer 2

1 + tan^2u = sec^2u is an identity so is that part of it?

Answer 3

@dpasingh ?

Answer 4

sorry some problem had arisen with my pc.

Answer 5

I am a little confused.. how can a triangle be used to verify?

Answer 6

sec x= AC/BC and tan x = AB/BC
i.e. \[\sec ^{2}x = [\frac{AC}{BC}] ^2 \]
\[\tan ^{2}x = [\frac{AB}{BC}] ^2 \]
\[\sec^2x-\tan^2x= \frac{AC^2}{BC^2}-\frac{AB^2}{BC^2}=\frac{AC^2-AB^2}{BC^2}=\frac{BC^2}{BC^2}=1\]
Hence
\[\sec^2x-\tan^2x= 1\]
Thus proved
Since by Pythagoras theorem: \[AC^2= AB^2+BC^2 \rightarrow AC^2-AB^2= BC^2\rightarrow BC^2=AC^2-AB^2\]

Answer 7

yes, 1 + tan^2x = sec^2 x
- tan^2x + 1 + tan^2x
= 1

Answer 8

Thank You guys! Really helpful!

Answer 9

yw