If sec2θ+tan2θ=1213, then sec4θ−tan4θ=.......... .

Acording to question,sec4θ−tan4θ=(sec2θ−tan2θ)(sec2θ+tan2θ) => 1×(sec2θ+tan2θ) ...........since, (sec2θ−tan2θ=1)=> 1213

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If ^{2}+^{2}={13}{12}, then ^{4}-^{4}=.......... .

Question

If $sec^{2} θ + tan^{2} θ = \frac{13}{12}$ , then $sec^{4} θ - tan^{4} θ =$ .......... .

\frac{12}{13}

\frac{13}{12}

1

\frac{1}{2}

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Henry Discussed

sec^{2} θ + tan^{2} θ = \frac{13}{12}

, then

sec^{4} θ - tan^{4} θ =

.......... .

13 mins ago

Discuss this question LIVE

13 mins ago

Acording to question,

se c^{4} θ - t a n^{4} θ = (se c^{2} θ - t a n^{2} θ) (se c^{2} θ + t a n^{2} θ)

1 \times (se c^{2} θ + t a n^{2} θ)

...........since,

(se c^{2} θ - t a n^{2} θ = 1)

\frac{13}{12}

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Question Text	If $sec^{2} θ + tan^{2} θ = \frac{13}{12}$ , then $sec^{4} θ - tan^{4} θ =$ .......... .
Answer Type	Text solution:1
Upvotes	150