The value of 4-16-43-16-62-16-4-16-2 is

4A = 1 tan^22A/2tan 2A ⇒1 = (1 tan^2A)^2 4tan^2A/4tan A(1 tan^2A), A = π/16 ⇒tan^4π/16+4tan^3π/16 6tan^2π/16 4tanπ/16+1 = 0 Put, tanπ/16 = 1/π/16 after replac ...

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Byju's Answer

Standard XII

Mathematics

Sin2A and Cos2A in Terms of tanA

The value of ...

Question

The value of ${cot}^{4} \frac{π}{16} - 4 {cot}^{3} \frac{π}{16} - 6 {cot}^{2} \frac{π}{16} + 4 cot \frac{π}{16} + 2$ is

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Solution

$cot 4 A = \frac{1 - {tan}^{2} 2 A}{2 tan 2 A} \Rightarrow 1 = \frac{(1 - {tan}^{2} A)^{2} - 4 {tan}^{2} A}{4 tan A (1 - {tan}^{2} A)}, A = \frac{π}{16} \Rightarrow {tan}^{4} \frac{π}{16} + 4 {tan}^{3} \frac{π}{16} - 6 {tan}^{2} \frac{π}{16} - 4 tan \frac{π}{16} + 1 = 0$
Put, $tan \frac{π}{16} = \frac{1}{cot \frac{π}{16}}$
after replacing it-
${cot}^{4} \frac{π}{16} - 4 {cot}^{3} \frac{π}{16} - 6 {cot}^{2} \frac{π}{16} + 4 {cot}^{4} \frac{π}{16} + 1 = 0$
${cot}^{4} \frac{π}{16} - 4 {cot}^{3} \frac{π}{16} - 6 {cot}^{2} \frac{π}{16} + 4 {cot}^{4} \frac{π}{16} + 2 = 1$

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The value of cot4π16−4cot3π16−6cot2π16+4cotπ16+2 is

cot4A = 1−tan22A2tan2A⇒1 = (1−tan2A)2−4tan2A4tanA(1−tan2A), A = π16⇒tan4π16+4tan3π16−6tan2π16−4tanπ16+1 = 0 Put, tanπ16 = 1cotπ16 after replacing it- cot4π16−4cot3π16−6cot2π16+4cot4π16+1 = 0 cot4π16−4cot3π16−6cot2π16+4cot4π16+2 = 1

The value of ${cot}^{4} \frac{π}{16} - 4 {cot}^{3} \frac{π}{16} - 6 {cot}^{2} \frac{π}{16} + 4 cot \frac{π}{16} + 2$ is