If sec(t) = 3, how do you find the exact value of $csc (90 - t)$ $csc(90 - t )$ ?

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Answer 1 · 2018-03-28T16:19:11

$sec (t) = 3$ $sec(t) = 3$
$\Rightarrow cos (t) = \frac{1}{3}$ $=> cos(t) = 1/3$

now, we know, $sin (90 - x) = cos x$ $sin(90-x) = cosx$

$sin (90 - t) = cos (t) = \frac{1}{3}$ $sin(90-t) = cos(t) = 1/3$

$csc (90 - t) = \frac{1}{sin (90 - t)} = 3$ $csc(90-t) = 1/sin(90-t) = 3$

If sec(t) = 3, how do you find the exact value of csc(90−t)csc(90 - t )?