Question

help me pls refer the picture attached for question please send solution and answer​

Answer 1

2 c0s 65 / sin ( 90_65) _ tan 20 / cot 30 _ sin 90

2cos 65/ cos 65 _ tan 20 / cot30 _ sin 90

2 _ tan20 / cot 30 _ sin 90 ...

ab answer k according tan cot ko sin cos mai break kro aur simple calculation sai answer pavo ....

Answer 2

Correct Question

• $\tt\dfrac{2cos65}{sin25}-\dfrac{Tan50}{Cot30}-sin90$

AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• $\tt\dfrac{2cos65}{sin25}-\dfrac{Tan50}{Cot30}-sin90$

${\sf{\green{\underline{\large{To\:find}}}}}$

• Value of this trignometic

${\sf{\pink{\underline{\Large{Explanation}}}}}$

• We find the value
• By help of complementary angle

$\tt\dfrac{2cos65}{sin25}-\dfrac{Tan50}{Cot30}-sin90$

• We know that
• cos(90-∅) = sin∅
• and Tan(90-∅) = cot∅

$\pink\rule{200}3$

$\implies\tt\dfrac{2cos(90-25)}{sin25}-\dfrac{Tan(90-50)}{Cot30}-sin90$

$\implies\tt\dfrac{2sin(25)}{sin25}-\dfrac{Cot(50)}{Cot30}-sin90$

$\implies\tt\cancel{\dfrac{2sin(25)}{sin25}}-\cancel{\dfrac{Cot(50)}{Cot30}}-sin90$

=>2-1+1

=>2

$\red\rule{200}3$

Hence value is 2