Evaluate the following - cos 37-sin 53-

We have, cos 37^∘sin 53^∘ = cos(90^∘ 53^∘)sin 53^∘ = sin 53^∘sin 53^∘ = 1 #160; #160; #160; [∵ cos(90^∘ θ) = sin ...

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Standard X

Mathematics

Complementary Trigonometric Ratios

Evaluate the ...

Question

Evaluate the following :
$\frac{c o s 37^{\circ}}{s i n 53^{\circ}}$

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Solution

We have,
$\frac{c o s 37^{\circ}}{s i n 53^{\circ}} = \frac{c o s (90^{\circ} - 53^{\circ})}{s i n 53^{\circ}} = \frac{s i n 53^{\circ}}{s i n 53^{\circ}} = 1$ $[∵ c o s (90^{\circ} - θ) = s i n θ]$

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(i) sin53° cos37° + cos53° sin37° = 1
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(iii) sec70° sin20° + cos20° cosec70° = 2
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Q. Without using trigonometric tables, prove that:
(i) sin53° cos37° + cos53° sin37° = 1
(ii) cos54° cos36° − sin54° sin36° = 0
(iii) sec70° sin20° + cos20° cosec70° = 2
(iv) tan 15° tan 60° tan 75° =

\sqrt{3}

(v) tan48° tan23° tan42° tan67° tan 45° = 1
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Evaluate the following :cos37∘sin53∘

We have,cos37∘sin53∘=cos(90∘−53∘)sin53∘=sin53∘sin53∘=1 [∵cos(90∘−θ)=sinθ]

Evaluate the following :
$\frac{c o s 37^{\circ}}{s i n 53^{\circ}}$

We have,
$\frac{c o s 37^{\circ}}{s i n 53^{\circ}} = \frac{c o s (90^{\circ} - 53^{\circ})}{s i n 53^{\circ}} = \frac{s i n 53^{\circ}}{s i n 53^{\circ}} = 1$ $[∵ c o s (90^{\circ} - θ) = s i n θ]$