If - is an acute angle and tan-2- then find the value of tan 7- 7-

We have, tan θ +cot θ =2 tan θ + 1/tan θ=2 (tan^2 θ +1)/tan θ=2 tan^2 θ 2 tan θ +1 =0 (tan θ 1)^2 =0 tan θ 1 =0 tan θ =1 tan θ =tan 45^∘ ⇒θ =45^∘ ⇒ tan ...

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Standard X

Mathematics

Trigonometric Identities

If θ is an ac...

Question

If $θ$ is an acute angle and $t a n θ + c o t θ = 2$ , then find the value of $t a n^{7} θ + c o t^{7} θ$ .

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Solution

We have,

$t a n θ + c o t θ = 2 t a n θ + \frac{1}{t a n θ} = 2 \frac{(t a n^{2} θ + 1)}{t a n θ} = 2 t a n^{2} θ - 2 t a n θ + 1 = 0 (t a n θ - 1)^{2} = 0 t a n θ - 1 = 0 t a n θ = 1 t a n θ = t a n 45^{\circ} \Rightarrow θ = 45^{\circ} \Rightarrow t a n^{7} θ + c o t^{7} θ = t a n^{7} 45^{\circ} + c o t^{7} 45^{\circ} = 1 + 1 = 2$

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If θ is an acute angle and tanθ+cotθ=2, then find the value of tan7θ+cot7θ.

We have, tanθ+cotθ=2tanθ+1tanθ=2(tan2θ+1)tanθ=2tan2θ−2tanθ+1=0(tanθ−1)2=0tanθ−1=0tanθ=1tanθ=tan45∘⇒θ=45∘⇒tan7θ+cot7θ=tan745∘+cot745∘=1+1=2

If $θ$ is an acute angle and $t a n θ + c o t θ = 2$ , then find the value of $t a n^{7} θ + c o t^{7} θ$ .