Prove that tan 4 -4 tan-1-tan 2-1-6 tan 2-tan 4-

We have LHS = tan 4θ = tan 2(2θ) =tan 2x, where x= 2θ =2tan x/1 tan^2x = 2tan 2θ/1 tan^2 2θ =2 ×2tan θ/(1 tan^2 θ)/1 ( 2tan θ/1 tan^2 θ )^2 = 4 tan θ(1 tan ^2 ...

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Standard VII

Mathematics

(a + b)^2 Expansion and Visualisation

Prove that ta...

Question

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

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Solution

We have

LHS = $tan 4 θ = tan 2 (2 θ)$

= $tan 2 x, where x = 2 θ$

= $\frac{2 t a n x}{1 - {tan}^{2} x} = \frac{2 tan 2 θ}{1 - {tan}^{2} 2 θ}$

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

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

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Similar questions

Q. Prove that

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

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θ \in (0, \frac{π}{2})

in which the given proof does not exist.

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

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Q. If

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\sec^{4} θ - \sec^{2} θ = \tan^{4} θ + \tan^{2} θ

Q. If

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show that

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Prove that tan 4θ=4tan θ(1−tan2θ)1−6 tan2θ+tan4θ

We have LHS = tan 4θ=tan 2(2θ) =tan 2x, where x= 2θ =2tan x1−tan2x=2tan 2θ1−tan2 2θ =2×2tan θ(1−tan2 θ)1−(2tan θ1−tan2θ)2=4 tan θ(1−tan 2θ)(1−tan2θ)2 −4tan2θ = 4tan θ(1−tan2θ)1−6tan2 θ+tan4 θ =RHS

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