Evaluate tan 45-cosec 30- - 60- 45- - 5 sin 90-2 0-

We have, ( tan 45^∘cosec 30^∘ )+ ( 60^∘ 45^∘ ) ( 5 sin 90^∘2 0^∘ )...(i) ` On substituting the trigonometric values we get, =12+21 52 =1+4 52=0 nbsp; ...

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Byju's Answer

Standard IX

Mathematics

Trigonometric Ratios of Complementary Angles

Evaluate tan ...

Question

Evaluate $(\frac{tan 45^{\circ}}{c o s e c 30^{\circ}}) + (\frac{sec 60^{\circ}}{cot 45^{\circ}}) - (\frac{5 sin 90^{\circ}}{2 cot 0^{\circ}})$

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Solution

We have,
$(\frac{tan 45^{\circ}}{c o s e c 30^{\circ}}) + (\frac{sec 60^{\circ}}{cot 45^{\circ}}) - (\frac{5 sin 90^{\circ}}{2 cot 0^{\circ}}) . . . (i)$
`
On substituting the trigonometric values we get,
$= \frac{1}{2} + \frac{2}{1} - \frac{5}{2} = \frac{1 + 4 - 5}{2} = 0$

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Evaluate (tan 45∘cosec 30∘)+(sec 60∘cot 45∘)−(5sin 90∘2cot 0∘)

We have, (tan 45∘cosec 30∘)+(sec 60∘cot 45∘)−(5sin 90∘2cot 0∘)...(i) ` On substituting the trigonometric values we get, =12+21−52=1+4−52=0

Evaluate $(\frac{tan 45^{\circ}}{c o s e c 30^{\circ}}) + (\frac{sec 60^{\circ}}{cot 45^{\circ}}) - (\frac{5 sin 90^{\circ}}{2 cot 0^{\circ}})$

We have,
$(\frac{tan 45^{\circ}}{c o s e c 30^{\circ}}) + (\frac{sec 60^{\circ}}{cot 45^{\circ}}) - (\frac{5 sin 90^{\circ}}{2 cot 0^{\circ}}) . . . (i)$
`
On substituting the trigonometric values we get,
$= \frac{1}{2} + \frac{2}{1} - \frac{5}{2} = \frac{1 + 4 - 5}{2} = 0$