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Grade 11Trigonometry

tanA=2x(x+1)/2x+1 then find the value of sinA and cosA,cosecA

Profile image of Siva
8 Years agoGrade 11
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1 Answer

Profile image of Sami Ullah
8 Years ago
tanA=\frac{2x^2+2x}{2x+1}
tanA=2[\frac{x^2+x}{2x+1}]
tan^{2}A=4[\frac{(x^2+x)^{2}}{(2x+1)^2}]
tan^{2}A=4[\frac{(x^4+x^2+2x^3)}{(4x^2+4x+1)}]
1+tan^{2}A=1+4[\frac{(x^4+x^2+2x^3)}{(4x^2+4x+1)}]
sec^{2}A=\frac{4x^2+4x+1+4(x^4+x^2+2x^3)}{4x^2+4x+1}
sec^{2}A=\frac{4x^4+8x^3+8x^2+4x+1}{4x^2+4x+1}
cos^{2}A=\frac{4x^2+4x+1}{4x^4+8x^3+8x^2+4x+1}
cosA=\frac{2x+1}{\sqrt{4x^4+8x^3+8x^2+4x+1}}
That is what I could do.I don’t think it can further be solved.Of it can be please notify me.Also you can now find sinA easily

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