Question

If tan theta +cot theta=3, prove that tan 4theta +cot 4 theta =47.​

Answer 1

Answer:

$tanθ+cotθ=3 \\ Squring \: on \: both \: the \: sides \\ {(tanθ + cotθ)}^{2} = {(3)}^{2} \\ {(tanθ)}^{2} + {(cotθ)}^{2} + 2(tanθ)(cotθ) = 9 \\ {tan}^{2} θ + {cot}^{2} θ + 2 \times tanθ \times \frac{1}{tanθ} = 9 \\ {tan}^{2} + {cot}^{2} θ + 2 = 9 \\ {tan}^{2} θ + {cot}^{2} θ = 9 - 2 \\ {tan}^{2}θ + {cot}^{2} θ = 7 \\ Squaring \: on \: both \: the \: sides \\ {( {tan}^{2} θ + {cot}^{2} θ)}^{2} = {(7)}^{2} \\ {( {tan}^{2}θ )}^{2} + { {(cot}^{2} θ)}^{2} + 2( {tan}^{2} θ)( {cot}^{2} θ) = 49 \\ {tan}^{4} θ + {cot}^{4} θ + 2 \times {tan}^{2} θ \times \frac{1}{ {tan}^{2}θ } = 49 \\ {tan}^{4} θ + {cot}^{4} θ + 2 = 49 \\ {tan}^{4} θ + {cot}^{4} θ = 49 - 2 \\ {tan}^{4} θ + {cot}^{4} θ = 47 \\ Hence \: proved$