Question

Evaluate 3 cos 68 . cosec 22 - 1/2 tan 43 . tan 47. tan 12 . tan 60 .tan78

Answer 1

3 cos 68. cosec 22 - 1/2 tan 43 . tan 47. tan 12. tan 60 . tan 78
⇒3 sin 22 . cosec 22 - 1/2 cot 47 .tan 47 .cot 78 . √3 . tan 78
(since cos 68 = sin 22, tan 43 = cot 47,tan 12 = cot 78 and tan 60 = √3)
⇒3 * 1/cosec 22 * cosec 22 - 1/2 cot 47 * 1/cot 47 . cot 78 . 1/cot 78 *√3
⇒3 - √3/2
⇒6 - √3    is the answer
       2

Answer 2

$3\cos 68 .\csc 22-\dfrac{1}{2}\tan 43 .\tan 47.\tan 12 .\tan 60 .\tan78=\dfrac{6-\sqrt{3}}{2}$

Step-by-step explanation:

We have,

$3\cos 68 .\csc 22-\dfrac{1}{2}\tan 43 .\tan 47.\tan 12 .\tan 60 .\tan78$

To find, $3\cos 68 .\csc 22-\dfrac{1}{2}\tan 43 .\tan 47.\tan 12 .\tan 60 .\tan78=?$

∴ $3\cos 68 .\csc 22-\dfrac{1}{2}\tan 43 .\tan 47.\tan 12 .\tan 60 .\tan78$

$=3\cos (90-22) .\csc 22-\dfrac{1}{2}\tan 43 .\tan (90-43).\tan 12 .\tan 60 .\tan (90-12)$

$=3\sin 22.\csc 22-\dfrac{1}{2}\tan 43 .\cot 43.\tan 12 .\tan 60 .\cot 12$

Using the trigonometric identity,

$\cos (90-22) =\sin A$ and $\tan (90-A)=\cot A$

$=3(1)-\dfrac{1}{2}(\tan 43 .\cot 43).(\tan 12 .\cot 12.)\tan 60$

Using the trigonometric identity,

$\sin A.\csc A=1$

$=3-\dfrac{1}{2}(1).(1).\tan 60$

Using the trigonometric identity,

$\tan A.\cot A=1$

$=3-\dfrac{1}{2}.\sqrt{3}$

$=\dfrac{6-\sqrt{3}}{2}$

Hence, $3\cos 68 .\csc 22-\dfrac{1}{2}\tan 43 .\tan 47.\tan 12 .\tan 60 .\tan78=\dfrac{6-\sqrt{3}}{2}$