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Grade 12Trigonometry

  1. tan(pi/4+cos inverse(a/b))+tan( pi/4-cos inverse (a/b))

Profile image of Jerry Joseph
7 Years agoGrade 12
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1 Answer

Profile image of Samyak Jain
7 Years ago
cos–1(a/b) = tan–1(\sqrt{b^2 - a^2} / a)
Use tan(A  + B) = (tanA – tanB)/(1 – tanAtanB) & tan(A  – B) = (tanA – tanB)/(1 + tanAtanB).
tan(\pi/4 + cos–1(a/b)) = [tan(\pi/4) + tan(cos–1(a/b))][1 – tan(\pi/4)tan(cos–1(a/b))]
                                  = [1 + tan(tan–1(\sqrt{b^2 - a^2} / a))][1 – tan(tan–1(\sqrt{b^2 - a^2} / a))]
                                  = (1 + \sqrt{b^2 - a^2} / a) / (1 – \sqrt{b^2 - a^2} / a)
                                  = (a + \sqrt{b^2 - a^2}) / (a – \sqrt{b^2 - a^2})
Similarly, tan(\pi/4 – cos–1(a/b))  =  (a – \sqrt{b^2 - a^2}) / (a + \sqrt{b^2 - a^2})
\therefore tan(\pi/4 + cos–1(a/b)) + tan(\pi/4 – cos–1(a/b))
     = [(a + \sqrt{b^2 - a^2}) / (a – \sqrt{b^2 - a^2})] + [(a – \sqrt{b^2 - a^2}) / (a + \sqrt{b^2 - a^2})]
     = [(a + \sqrt{b^2 - a^2})2 + (a – \sqrt{b^2 - a^2})2] / [(a – \sqrt{b^2 - a^2})(a + \sqrt{b^2 - a^2})]
     = [2a2 + 2(b2 – a2)] / [a2 – (b2 – a2)]
     = 2 b2/(2a2 – b2) 

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