Thread: Trigonometry

a) Sketch the graphs of y= tanA and y= secA

b) Prove that secA- tanA= 1 / (secA+tanA)

c) Deduce from parts a) and b) that secA-tanA is greater than 0 and less and equal to 1,

for values of A such that A is greater and equal to 0 and less than and equal to 90 degrees. Explain your reasoning clearly.

d) Given that secA-tanA= 1/2

find the exact value of cosA

Originally Posted by Oasis1993

a) Sketch the graphs of y= tanA and y= secA

b) Prove that secA- tanA= 1 / (secA+tanA)

c) Deduce from parts a) and b) that secA-tanA is greater than 0 and less and equal to 1,

for values of A such that A is greater and equal to 0 and less than and equal to 90 degrees. Explain your reasoning clearly.

d) Given that secA-tanA= 1/2

find the exact value of cosA

HI

(b)









some cancellings ..

(d)





*

where and



evaluate r and alpha respectively and put it into the equation * .

Originally Posted by Oasis1993

a) Sketch the graphs of y= tanA and y= secA

b) Prove that secA- tanA= 1 / (secA+tanA)

c) Deduce from parts a) and b) that secA-tanA is greater than 0 and less and equal to 1,

for values of A such that A is greater and equal to 0 and less than and equal to 90 degrees. Explain your reasoning clearly.

d) Given that secA-tanA= 1/2

find the exact value of cosA

HI

from (b) , secA- tanA= 1 / (secA+tanA)





When

is max , its denominator would be minimum ie sin A=0 , correspondingly cos A=1




Similarly , when its min , denominator would be max , ie sin A=1, cos A=0 ,



so we can see that