SOLUTION: Respected Sir, I am really amazed to see such a wonderful work on net, which becomes a very useful guide for the students. I am having some problem with the subject 'Trigonometry

Question 184174: Respected Sir,
I am really amazed to see such a wonderful work on net, which becomes a very useful guide for the students.
I am having some problem with the subject 'Trigonometry'. Here are the question.
1) sec^2@-sin^2@.sec^2@=1
2) (1+2sin^2@)/(1+3tan^2@)=cos^2@
I hope for a quick reply

Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
1) sec^2@-sin^2@.sec^2@=1
sec^2@(1-sin^2@)=1
but 1-sin^2@=cos^2@ since sin^2@+cos^2@=1 (fundamental relationship)
And
sec^2@=1/cos^2@ since sec@= 1/cos@ (fundamental relationship)
So
cos^2@/cos^2@=1
1=1
2) (1+2sin^2@)/(1+3tan^2@)=cos^2@
NUMERATOR: substitute sin^2@=1-cos^2@ (fundamental relationship)
(1+2(1-cos^2@))/(1+3tan^2@)=
(1+2-2cos^2@)/(1+3tan^2@)=
(3-2cos^2@)/(1+3tan^2@)=
DENOMINATOR:
substitute tan^2@=sin^2@/cos^2@ (tan@= sin@/cos@ = fundamental relationship)
(3-2cos^2@)/(1+3(sin^2@/cos^2@)=
(3-2cos^2@)/(cos^2@+3sin^2@)/cos^2@)=
(3-2cos^2@)/(cos^2@+3sin^2@)/cos^2@)=
substitute sin^2@=1-cos^2@ (fundamental relationship)
(3-2cos^2@)/(cos^2@+3(1-cos2@)/cos^2@)=
(3-2cos^2@)/(cos^2@+3-3cos2@)/cos^2@)=
(3-2cos^2@)/(3-2cos^2@)/cos^2@)=
FINISH SIMPLIFYING THE FRACTION 3-2cos^2@ in numerator and denominator cancel
cos^2@(3-2cos^2@)/(3-2cos2@)=cos^2@
Hope this helps---ptaylor

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