Prove the following identities
(a)1-sin A / 1 + sin A = (sec A - tan A)^2
(b)2 cos x = 3 tan x
For the first question, I have been trying to do it using the sin^2 + cos^2 =1 method... Or you can say I have almost looked through all the trigo formulas. Somehow, nothing seems to be able to fit the equations.
Then for the second question I did some shifting of the terms but to no avail.
I would appreciate anyone and everyone's help. However, I would go attempt the questions again before I come to take a look at the solutions if I really can't solve them.
Thanks a lot. =))))))))))))))
for question (a):
(sec A - tan A)^2
expand first using (a + b)^2 = a^2 - 2ab + b^2.
= (sec A)^2 - 2 sec A tan A + (tan A)^2
expand further.
= (1 / cos^2 A) - [2 (1 / cos A)(sin A / cosA) ] + ( sin^2 A / cos^2 A)
expand further and try to simplify.
= (1 - 2 sin A + sin^2 A) / (cos^2 A)
this is what you get after simplifying the above chunk of stuffs. remember sin^2 + cos^2 =1 and hence, change the denominator.
= (1 - 2 sin A + sin^2 A) / (1 - sin^2 A)
change the numerator from a^2 - 2ab + b^2 form to (a + b)^2 form. change the denominator from a^2 - b^2 form to (a + b)(a - b).
= (1 - sin A)^2 / [ (1 + sin A)(1 - sin A) ]
you can choose to open up the numerator like what i've done below. this is optional lah.
= [ (1 - sin A)(1 - sin A) ] / [ ( 1 + sin A)(1 - sin A) ]
cancel the common factor which is (1 - sin A) in this case.
= 1 - sin A / 1 + sin A (proven)
hope this helps. =)
part (b) is not an identity.
In general, you cannot convert cos to tan just like that,
i.e. cos x = 3 tan x / 2