Question

Evaluate :
[3sin 43°/cos 47°]^2 - cos 37° cosec53°/tan 5° tan 25° tan 45° tan 65° tan 85° .

Give this solution with step by step explanation.​

Answer 1

Step-by-step explanation:

the solution is in the above image.

if you cannot understand this then ask me in the comment

Answer 2

$(\dfrac{3\sin 43}{\cos 47})^2-\dfrac{\cos 37 \csc 53}{\tan 5 \tan 25 \tan 45 \tan 65 \tan 85}$ = 8

Step-by-step explanation:

We have,

$(\dfrac{3\sin 43}{\cos 47})^2-\dfrac{\cos 37 \csc 53}{\tan 5 \tan 25 \tan 45 \tan 65 \tan 85}$

To find, $(\dfrac{3\sin 43}{\cos 47})^2-\dfrac{\cos 37 \csc 53}{\tan 5 \tan 25 \tan 45 \tan 65 \tan 85}$ = ?

∴ $(\dfrac{3\sin 43}{\cos 47})^2-\dfrac{\cos 37 \csc 53}{\tan 5 \tan 25 \tan 45 \tan 65 \tan 85}$

= $(\dfrac{3\sin 43}{\cos (90-43)})^2-\dfrac{\cos 37 \csc (90-37)}{\tan 5 \tan 25 \tan 45 \tan (90-15) \tan (90-5)}$

Using the trigonometric identity,

$\sin A=\cos (90-A), \sec A=\csc (90-A)$ and $\cot A =\tan (90-A)$

$=(\dfrac{3\sin 43}{\sin 43})^2-\dfrac{\cos 37 \sec 37}{\tan 5 \tan 25 (1) \cot 15 \cot 15}$

Using the trigonometric identity,

$\cos A \sec A=1$ and $\cos A \sec A=1$

$=(3)^2-\dfrac{1}{1}$

= 9 - 1

= 8

∴  $(\dfrac{3\sin 43}{\cos 47})^2-\dfrac{\cos 37 \csc 53}{\tan 5 \tan 25 \tan 45 \tan 65 \tan 85}$ = 8