Find the value of sin-50o-cos-40o-cosec-40o-sec-50o - 4 cos 50o cosec 40o-

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Question

Find the value of $\frac{s i n 50^{o}}{c o s 40^{o}} + \frac{c o s e c 40^{o}}{s e c 50^{o}} - 4$ cos $50^{o}$ cosec $40^{o}$ .

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Solution

As we know

$s i n 50^{o} = s i n (90 - 40) = c o s 50^{o}$

$c o s e c 40^{o} = c o s e c (90 - 50) = s e c 40^{o}$

$c o s 50^{o} = c o s (90 - 40) = s i n 40^{o}$

$s i n 40^{o} c o s e c 40^{o} = 1$

now put these values in this,

$\frac{s i n 50^{o}}{c o s 40^{o}} + \frac{c o s e c 40^{o}}{s e c 50^{o}} - 4$ cos $50^{o}$ cosec $40^{o}$

and we get,

$fraction numerator cos 40 over denominator cos 40 end fraction plus fraction numerator s e c 50 over denominator s e c 50 end fraction minus 4 sin 40 cos e c 40 1 plus 1 minus 4 equals negative 2$

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Find the value of sin 50ocos 40o+cosec 40osec 50o−4 cos 50o cosec 40o.

As we know sin50o=sin(90−40)=cos50o cosec40o=cosec(90−50)=sec40o cos50o=cos(90−40)=sin40o sin40o cosec40o=1 now put these values in this, sin 50ocos 40o+cosec 40osec 50o−4 cos 50o cosec 40o and we get,

Find the value of $\frac{s i n 50^{o}}{c o s 40^{o}} + \frac{c o s e c 40^{o}}{s e c 50^{o}} - 4$ cos $50^{o}$ cosec $40^{o}$ .