Question

sin 18/cos 72+√3(tan10 tan30 tan40 tan50 tan80),Evaluate it.

Answer 1

= sin18°/cos72° + √3(tan10°.tan30°.tan40°.tan50°.tan80°)

we know, sin(90° - x) = cosx => sin18° = sin(90° - 72°) = cos72° .
hence, sin18° = cos72° -----(1)

again, tan(90° - x) = cotx
so, tan10° = tan(90° - 80°) = cot80° ----(2)
tan40° = tan(90° - 50°) = cot50° -----(3)

now, From equations (1), (2) and (3),
= cos72°/cos72° + √3(cot80°.tan30°.cot50°.tan50°.tan80°)

= 1 + √3{(tan80°.cot80°) × (tan50°.cot50°) × tan30°}

= 1 + √3 × tan30°

= 1 + √3 × 1/√3

= 1 + 1 = 2 [ans]

Answer 2

We know that:

(i) sin(90 - A) = cosA.

(ii) cotA = (1/tanA).

(iii) tan(90 - A) = cotA

Now, Coming to the question.

$= > \frac{sin 18}{cos 72} + \sqrt{3}(tan 10 * tan 30 * tan 40 * tan 50 * tan 80)$

$= > \frac{sin(90 - 72)}{`cos 72} + \sqrt{3}(tan 10 * \frac{1}{\sqrt{3}} * tan(90 - 50) * tan 50 * tan(90 - 10))$

$= > \frac{cos 72}{cos 72} + \sqrt{3}(tan 10 * \frac{1}{\sqrt{3}} * cot 50 * tan 50 * cot 10)$

$= > 1 + \sqrt{3}(tan 10 * \frac{1}{\sqrt{3}} * \frac{1}{tan 50} * tan 50 * \frac{1}{tan 10} )$

$= > 1 + \sqrt{3}(\frac{1}{\sqrt{3}})$

= > 1 + 1

= > 2.

Hope this helps!