1. ## 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

2. 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) $\displaystyle \frac{1}{\sec A+\tan A}$

$\displaystyle \frac{\cos A}{1+\sin A}$

$\displaystyle \frac{\cos A}{1+\sin A}\cdot \frac{1-\sin A}{1-\sin A}$

$\displaystyle \frac{\cos A(1-\sin A)}{1-\sin^2 A}$

$\displaystyle \frac{\cos A(1-\sin A)}{\cos^2 A}$

some cancellings .. $\displaystyle \sec A -\tan A$

(d) $\displaystyle \sec A-\tan A=\frac{1}{2}$

$\displaystyle \frac{1-\sin A}{\cos A}=\frac{1}{2}$

$\displaystyle \cos A+2\sin A=2$

$\displaystyle r\cos (A-\alpha)=2$ *

where $\displaystyle r=\sqrt{1^2+2^2}$ and

$\displaystyle \tan \alpha=2\Rightarrow \alpha=\arctan 2$

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

3. 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)

$\displaystyle \frac{1}{\sec A+\tan A}$

$\displaystyle \frac{\cos A}{1+\sin A}$

When $\displaystyle \frac{\cos A}{1+\sin A}$

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

$\displaystyle \frac{1}{1+0}=1$

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

$\displaystyle \frac{0}{2}=0$

so we can see that $\displaystyle 0\leq \sec A-\tan A\leq 1$