InterviewSolution
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InterviewSolution
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\begin{array}{r}{\tan ^{2} \frac{\pi}{16}+\tan ^{2} \frac{2 \pi}{16}+\tan ^{2} \frac{3 \pi}{16}+\tan ^{2} \frac{4 \pi}{16}+\tan ^{2} \frac{5 \pi}{16}} \\ {+\tan ^{2} \frac{6 \pi}{16}+\tan ^{2} \frac{7 \pi}{16}=35}\end{array} |
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Answer» tan²(π/16) + tan²(7π/16)= [tan(π/16) + tan(7π/16)]²−2 tan(π/16)tan(7π/16)= [tan(π/16) + tan(7π/16)]² −2Similarly,tan²(3π/16)+ tan²(5π/16)= [tan(3π/16) + tan(5π/16)]² −2tan²(2π/16)+ tan²(6π/16)= [tan(2π/16) + tan(6π/16)]² −2 Now, [tan(π/16) + tan(7π/16)]²= [sin(7π/16 +π/16) / ((cos(π/16) cos(7π/16))]²= 1/((sin(7π/16) cos(7π/16))²= 4/sin²(π/8) Similarly,[tan(2π/16) + tan(6π/16)]²= 4/sin²(π/4)[tan(3π/16) + tan(5π/16)]²= 4/sin²(6π/16) Therefore,tan²(π/16) +tan²(2π/16) +tan²(3π/16) +tan²(4π/16) +tan²(5π/16) +tan²(6π/16) +tan²(7π/16)= 4/sin²(π/8) −2 + 4/sin²(π/4) −2 + 4/sin²(6π/16) −2 + tan²π/4= 4/sin²(π/8) + 4/sin²(6π/16) + 3= 4[1/sin²(π/8) + 1/sin²(3π/8)] + 3= 4[sin²(π/8) + cos²(π/8] / [sin²(π/8) cos²(π/8)] + 3= 16/sin²(π/4) + 3=35 |
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