Prove that tan 4- -4 tan- 1-tan2-1-6tan2-tan4- Find the angle -0-2 in which the given proof does not exist-

tan 4θ =tan (2× 2θ ) =2tan (2θ)1 tan ^2 (2θ ) =2 [ 2tanθ1 tan ^2θ ]1 [ 2 tanθ1 tan ^2θ ]^2 =4tanθ1 tan ^2θ(1 tan ^2θ)^2 4tan ^2θ(1 tan ...

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Standard XII

Mathematics

Algebra of Complex Numbers

Prove that ...

Question

Prove that $tan 4 θ = \frac{4 tan θ (1 - {tan}^{2} θ)}{1 - 6 {tan}^{2} θ + {tan}^{4} θ}$ . Find the angle $θ \in (0, \frac{π}{2})$ in which the given proof does not exist.

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Solution

$tan 4 θ$

$= tan (2 \times 2 θ)$

$= \frac{2 tan (2 θ)}{1 - {tan}^{2} (2 θ)}$

$= \frac{2 [\frac{2 tan θ}{1 - {tan}^{2} θ}]}{1 - {[\frac{2 tan θ}{1 - {tan}^{2} θ}]}^{2}}$

$= \frac{\frac{4 tan θ}{1 - {tan}^{2} θ}}{\frac{(1 - {tan}^{2} θ)^{2} - 4 {tan}^{2} θ}{(1 - {tan}^{2} θ)^{2}}}$

$= \frac{4 tan θ (1 - {tan}^{2} θ)}{(1 - {tan}^{2} θ)^{2} - 4 {tan}^{2} θ}$

$= \frac{4 tan θ (1 - {tan}^{2} θ)}{1 + {tan}^{4} θ - 2 {tan}^{2} θ - 4 {tan}^{2} θ}$

$= \frac{4 tan θ (1 - {tan}^{2} θ)}{1 - 6 {tan}^{2} θ + {tan}^{4} θ}$

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Prove that tan4θ=4tanθ(1−tan2θ)1−6tan2θ+tan4θ. Find the angle θ∈(0,π2) in which the given proof does not exist.

tan4θ=tan(2×2θ)=2tan(2θ)1−tan2(2θ)=2[2tanθ1−tan2θ]1−[2tanθ1−tan2θ]2=4tanθ1−tan2θ(1−tan2θ)2−4tan2θ(1−tan2θ)2=4tanθ(1−tan2θ)(1−tan2θ)2−4tan2θ=4tanθ(1−tan2θ)1+tan4θ−2tan2θ−4tan2θ=4tanθ(1−tan2θ)1−6tan2θ+tan4θ

Prove that $tan 4 θ = \frac{4 tan θ (1 - {tan}^{2} θ)}{1 - 6 {tan}^{2} θ + {tan}^{4} θ}$ . Find the angle $θ \in (0, \frac{π}{2})$ in which the given proof does not exist.