Hi viewers, this is my third post on Trig Identities, I am fed up with this topic, anyways lets start, many still complain that they still face challenges with trig identities, so I pick some difficult(complex) questions for us to solve. Remember if you haven't read my previous posts try to check them out.
Now lets begin
1. Prove 1+cosx/1-cosx ➕ 1-cosx/1+cosx = 4cot²x +2
Solution.
When u see a fraction you try your best to get a common denominator which in this case we need to find the lowest common multiple (LCM). When there is nothing common in your denominators;the lcm is the product of the denominator.
1. If sinx=3/5, find the values of cosx, tanx, cotx and cosecx.
Solution.
From SOHCAHTOA,
Sinx=opp/hyp=3/5
using Pythagoras theorem to find the adjacent we have:
5²=3²+adj²
adj²=5²-3²=25-9=16
adj= √16=4.
cosx=adj/hyp=4/5
tanx=opp/adj=3/4
cotx=1/tanx=1/3/4=4/3
cosecx=1/sinx=1/3/5=5/3.
Now let us start proving of Trig identities.
1. Prove (1+tan²x)(1-sin²x) = 1
Solution.
If you remember the seven trig identities in my previous post Introduction to proving Trig identities You can read that post also.
In that post I prove 1+tan²x=sec²x
1-sin²x=cos²x.
Substituting we have
(sec²x)×(cos²x)
but sec²x=1/cos²x,Substituting we have,
1/cos²x ×cos²x = 1. (proved).
2. Prove cos²x-sin²x=2cos²x - 1.
Now we plan on getting 2cos²x - 1 as our answer, so we will manipulate our sin²x to get cos²x.
sin²x=1-cos²x. Substituting we have,
cos²x-(1-cos²x)
cos²x -1+cos²x
collect like terms
cos²x+cos²x-1
2cos²x - 1 (proved).
3. Prove (1-sinx)(1+sinx)/sin²x = cot²x.
Solution.
Expanding the numerator we have
1-sin²x/sin²x.
but 1-sin²x=cos²x, hence we have,
cos²x/sin²x =cot²x (proved).
This post is proving basic trig identities, by Monday I will release proving of complex Trig identities.
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