Home › Forums › Math questions › Geometry questions

proving trigonometry

Posted July 4th, 2010 by aloe
in

tan # + sec # -1
--------------------- = tan # +sec #
tan # - sec # + 1

# = teta

$$\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}=\tan \theta+\sec \theta$$
‹ COORDINATE GEOMETRY- PARABOLAS WHOSE AXES ARE PARALLEL TO EITHER X-AXIS OR Y-AXIS FIND THE VOLUME OF THE CLAY REQUIRED... ›

Comments

answer

On July 5th, 2010 aloe says:

tan x + sec x -1
= --------------------
tan x - sec x + 1

tan x + sec x - ( sec^2 (x) - tan ^ 2 (x) )
= ----------------------------------------------------
tan x - sec x + 1

tan x + sec x - ( sec x - tan x)(sec x + tan x)
= -------------------------------------------------------
tan x - sec x + 1

tan x + sec x (1 - (sec x - tan x))
= ------------------------------------------
tan x - sec x + 1

= tan x + sec x

answer

On July 5th, 2010 aloe says:

tan x + sec x -1
= --------------------
tan x - sec x + 1

tan x + sec x - ( sec^2 (x) - tan ^ 2 (x) )
= ----------------------------------------------------
tan x - sec x + 1

tan x + sec x - ( sec x - tan x)(sec x + tan x)
= -------------------------------------------------------
tan x - sec x + 1

tan x + sec x (1 - (sec x - tan x))
= ------------------------------------------
tan x - sec x + 1

= tan x + sec x

please

On July 4th, 2010 aloe says:

please help me...i need d answer as soon as possible... please...
we need 2 discuss it 2day morning....

answer

On July 5th, 2010 Structure says:
$$\frac{\tan x+\sec x-1}{\tan x-\sec x+1}=\tan x+\sec x$$

is equivalent to

$$\tan x+\sec x-1=(\tan x-\sec x+1)(\tan x+\sec x)$$

or

$$\tan x+\sec x-1=\tan ^2x-\sec^2 x+\tan x+\sec x$$

or

$$\sec^2x=1+\tan^2 x$$

other solution

On July 5th, 2010 Structure says:
$$\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}=\frac{\frac{\sin\theta}{\cos \theta}+\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}-\frac{1}{\cos \theta}+1}=\frac{\sin\theta+\cos\theta-1}{\sin\theta-\cos\theta+1}=\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}+2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}-2\sin^2\frac{\theta}{2}}=\frac{\cos\frac{\theta}{2}+\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}=$$
$$=\frac{(\cos\frac{\theta}{2}+\sin\frac{\theta}{2})^2}{\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}}=\frac{1+\sin\theta}{\cos\theta}=\tan \theta+\sec \theta$$

Back to top