Question

1 tan 50° + sec 50
$\div$
cot 40+cosec 40°​

Answer 1

Answer:

Value of \frac{(tan50+sec50)}{(cot40+cosec40)}+cos40\times cosec50 < /p > < p > =2

(cot40+cosec40)

(tan50+sec50)

+cos40×cosec50</p><p>=2

Step-by-step explanation:

Given

\frac{(tan50+sec50)}{(cot40+cosec40)}+cos40\times cosec50

(cot40+cosec40)

(tan50+sec50)

+cos40×cosec50

=\frac{tan(90-40)+sec(90-40)}{(cot40+cosec40)}+cos40\times cosec(90-40)

(cot40+cosec40)

tan(90−40)+sec(90−40)

+cos40×cosec(90−40)

= \frac{(cot40+cosec40)}{(cot40+cosec40)}+cos40\times sec40

(cot40+cosec40)

(cot40+cosec40)

+cos40×sec40

=1+cos40 \times \frac{1}{cos40}1+cos40×

cos40

1

= 1+11+1

=22

Therefore,

Value of \frac{(tan50+sec50)}{(cot40+cosec40)}+cos40\times cosec50=2

(cot40+cosec40)

(tan50+sec50)

+cos40×cosec50=2