If sin-35- where -0-2-then sec2-tan2- is 17k-then the value of k is -

Given nbsp;sinθ=35=(opposite)(hypotenuse)⇒adjacent nbsp;side=52 32=4. Now, nbsp;we can write nbsp; nbsp; secθ=54 nbsp;and nbsp;tanθ=34. sec2θ+tan2 ...

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Standard XII

Mathematics

Trigonometric Ratios Using Right Angled Triangle

If sinθ=35, w...

Question

$If sinθ = \frac{3}{5}, where θ \in (0, \frac{π}{2}), then ({sec}^{2} θ + {tan}^{2} θ) is 17 k .$
then the value of k is .

\frac{1}{8}

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\frac{1}{16}

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\frac{1}{32}

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Solution

The correct option is A $\frac{1}{8}$
Given $sin θ = \frac{3}{5} = \frac{(opposite)}{(hypotenuse)} \Rightarrow adjacent side = \sqrt{5^{2} - 3^{2}} = 4 .$
Now, we can write
$s e c θ = \frac{5}{4} and tan θ = \frac{3}{4}$ .
$s e c^{2} θ + {tan}^{2} θ = \frac{25}{16} + \frac{9}{16} \Rightarrow s e c^{2} θ + {tan}^{2} θ = \frac{34}{16} = \frac{17}{8} .$

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If sinθ=35, where θ∈(0,π2),then (sec2θ+tan2θ) is 17k. then the value of k is .

The correct option is A 18Given sinθ=35=(opposite)(hypotenuse)⇒adjacent side=√52−32=4. Now, we can write secθ=54 and tanθ=34. sec2θ+tan2θ=2516+916⇒sec2θ+tan2θ=3416=178.

$If sinθ = \frac{3}{5}, where θ \in (0, \frac{π}{2}), then ({sec}^{2} θ + {tan}^{2} θ) is 17 k .$
then the value of k is .