Question

sin2 25+sin2+65+√3(tan 5 tan 15 tan 30 tan 75tan 85

Answer 1

Sin²25°+sin²65°+√3(tan5°tan15°tan30°tan75°tan85°)
=sin²25°+sin²(90°-25°)+√3{tan5°tan15°×(1/√3)×tan(90°-15°)tan(90°-5°)}
=sin²25°+cos²25°+√3×1/√3(tan5°tan15°cot15°cot5°)
=1+1
=2 Ans.

Answer 2

$\sin^225+\sin^265+\sqrt{3}(\tan 5\tan 15\tan 30\ tan75\tan 85)=2$

Step-by-step explanation:

We have,

$\sin^225+\sin^265+\sqrt{3}(\tan 5\tan 15\tan 30\ tan75\tan 85)$

To find,$\sin^225+\sin^265+\sqrt{3}(\tan 5\tan 15\tan 30\ tan75\tan 85)=?$

∴ $\sin^225+\sin^265+\sqrt{3}(\tan 5\tan 15\tan 30\ tan75\tan 85)$

=$\sin^225+\sin^2(90-25)+\sqrt{3}(\tan 5\tan 15\tan 30\tan(90-15)\tan (90-5))$

Using trigonometric identity,

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

$=\sin^225+\cos^2 25+\sqrt{3}(\tan 5\tan 15\tan 30\cot 15\cot 5)$

$=1+\sqrt{3}((\tan 5.\cot 5).(\tan 15.\cot 15)\tan 30)$

Using trigonometric identity,

$\sin^2A+\cos^2 A=1$

$=1+\sqrt{3}((1).(1)\tan 30)$

Using trigonometric identity,

$\tan A.\cot A=1$

$=1+\sqrt{3}.\tan 30$

$=1+\sqrt{3}.\dfrac{1}{\sqrt{3}}$

= 1 + 1

= 2

Hence, $\sin^225+\sin^265+\sqrt{3}(\tan 5\tan 15\tan 30\ tan75\tan 85)=2$