Please simplify this trigonometric question for me
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Step-by-step explanation:
tanA+secA-1)(tanA+1+secA)=2sinA/1-sinA
secA=1/cosA
secA-1=(1-cosA)/cosA
tanA+secA-1=(sinA-cosA+1)/cosA
(tanA+1+secA)=(sinA+cosA+1)/cosA
multiply
[(sinA-cosA+1)/cosA][(sinA+cosA+1)/cosA...
[sin^2A+sinAcosA+sinA-sinAcosA-cos^2A-c...
=[sin^2A-cos^2A+2sinA+1]/cos^2A
we have sin^2A+cos^2A=1=>1-cos^2A=sin^2A
=>[2sin^2A+2sinA]/cos^2A
and cos^2a=1-sin^2A=(1+sinA)(1-sinA)
2sin^2A+2sinA can be written as: (1+sinA)(2sinA)
=>(1+sinA)(2sinA)/(1+sinA)(1-sinA)
=2sinA/1-sinA
kritinmishra12357:
What you had done in 4th step that is tanA+secA-1=(sinA-cosA+1)/cosA
It's correct ! I never do wrong
Ya , I am not saying that you are wrong but I didn't understand that step .please elaborate
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