Find the value of the following-i sin 50-cos 40-cosec 40- 50-4 cos 50-cosec 40-

Sin #160;50 #176;cos #160;40 #176;+cosec #160;40 #176;sec #160;50 #176; 4cos #160;50 #176;.cosec #160;40 #176;= #160;sin #160;(90 40 ...

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Standard IX

Mathematics

Introduction to Trigonometry

Find the valu...

Question

Find the value of the following.
(i) $\frac{\sin 50 °}{\cos 40 °} + \frac{\cos e c 40 °}{s e c 50 °} - 4 \cos 50 ° . \cos e c 40 °$

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Solution

$\frac{\sin 50 °}{\cos 40 °} + \frac{cosec 40 °}{\sec 50 °} - 4 \cos 50 ° . cosec 40 ° = \frac{\sin (90 - 40) °}{\cos 40 °} + \frac{cosec (90 - 50) °}{\sec 50 °} - 4 \cos (90 - 40) ° cosec 40 ° = \frac{\cos 40 °}{\cos 40 °} + \frac{\sec 50 °}{\sec 50 °} - 4 \sin 40 ° cosec 40 ° [∵ cosec (90 - θ) = \sec θ] & [\sin (90 - θ) = \cos θ] = 1 + 1 - 4 = - 2$

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(ii) $4 (s i n^{4} 30^{o} + c o s^{4} 60^{o}) - \frac{2}{3} (s i n^{2} 60^{o} - c o s^{2} 45^{o}) + \frac{1}{2} t a n^{2} 60^{o}$

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(vi) $t a n 7^{o} t a n 23^{o} t a n 60^{o} t a n 67^{o} t a n 83^{o}$

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(x) $\frac{c o s 58^{o}}{s i n 32^{o}} + \frac{s i n 22^{o}}{c o s 68^{o}} - \frac{c o s 38^{o} c o s e c 52^{o}}{t a n 18^{o} t a n 35^{o} t a n 60^{o} t a n 72^{o} t a n 55^{o}}$

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Find the value of the following. (i) sin50°cos40° + cosec40°sec50° - 4cos 50°. cosec 40°

sin 50°cos 40°+cosec 40°sec 50°-4cos 50°.cosec 40°= sin (90-40)°cos 40°+cosec (90-50)°sec 50°-4cos (90-40)°cosec 40°= cos 40°cos 40°+ sec 50°sec 50°-4sin 40°cosec 40° ∵cosec90-θ=sec θ & sin90-θ=cos θ= 1+1-4= -2

Find the value of the following.
(i) $\frac{\sin 50 °}{\cos 40 °} + \frac{\cos e c 40 °}{s e c 50 °} - 4 \cos 50 ° . \cos e c 40 °$