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prove this?
1. (sinA+cosecA)*(sinA+cosecA) +(cosA+secA)*(cosA+secA)=tanA*tanA = cotA*cotA+7
2. (1+cotA-cosecA)(1+tanA+secA)=2
3. (cosecA-sinA)(secA-cosA)(tanA+cotA)=1
4. tanA+secA-1/tanA-secA+1=1+sinA/cosA
5. tanA+secA-1/tanA-secA+1=tanA*secA
please prove
8 Answers
- 1 decade agoFavorite Answer
i don't know if i'll have the patience to do all of these, but it's really just definition chasing. you should be able to do these
1) (sinA+cosecA)*(sinA+cosecA) +(cosA+secA)*(cosA+secA)
= sin^2(A) + 2 + csc^2(A) + cos^2(A) + 2 + sec^2(A)
= 5 + csc^2(A) + sec^2(A)
= 6 + csc^2(A) + tan^2(A)
= 7 + cot^2(A) + tan^2(A)
(i think that's what you meant it to equal)
2) (1+cotA-cosecA)(1+tanA+secA)
= 1 + cotA - cscA + tanA + 1 - secA + secA + cscA - cscAsecA
= 2 + cot(A) + tan(A) - 1/(sin(A)cos(A))
= 2 + cos^2(A)/(sin(A)cos(A)) + sin^2(A)/(sin(A)cos(A)) - 1/(sin(A)cos(A))
= 2 + 1/(sin(A)cos(A)) - 1/(sin(A)cos(A))
= 2
you didn't finish writing the problem for 3 through 5.
- gnegyLv 44 years ago
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- 1 decade ago
hey,,,,i knwo yahoo answers is wonderful...but that does not mean u have to take advantage of it...i am a maths teacher...and it would be great if u try to prove urself. after all they are not that difficult......it will be good for you....
- Anonymous1 decade ago
its so easy you should do yourself or you will fail in the exam for sure. why are u so lazy?
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- Anonymous1 decade ago
Ask your math teacher.
- 1 decade ago
aahaa...........i suppose you are so......so.........so........intelligent..........