AP EAPCET 18 May 2024 Shift 1 Solved Paper

To solve the given expression tan‌6∘+tan‌42∘+tan‌66∘+tan‌78∘, we can apply some identities and symmetry properties of the tangent function.
First, let's use the identity for tangent of a sum and difference of angles:
tan(A+B)‌tan(A−B)=(‌

tan‌A+tan‌B

1−tan‌A‌tan‌B

)(‌

tan‌A−tan‌B

1+tan‌A‌tan‌B

)
This can be simplified to:
tan(A+B)‌tan(A−B)=‌

tan2A−tan2B

1−tan2Atan2B

Now, substitute A=60∘ and B=18∘ into equation (i):
tan‌78∘×tan‌42∘=‌

tan260∘−tan218∘

1−tan260∘tan218∘

=‌

3−tan218∘

1−3tan218∘

Simplifying:
=‌

1

tan‌18∘

[‌

3‌tan‌18∘−tan318∘

1−3tan218∘

]
So,
tan‌78∘‌tan‌42∘=‌

tan‌54∘

tan‌18∘

Next, substitute A=60∘ and B=54∘ in equation (i):
tan‌114∘×tan‌6∘=‌

3−tan254∘

1−3tan254∘

=‌

1

tan‌54∘

[‌

3‌tan‌54∘−tan354∘

1−3tan254∘

]
This gives:
=‌

tan‌162∘

tan‌54∘

Since tan‌114∘=tan(180∘−66∘)=−tan‌66∘ and tan‌162∘=tan(180∘−18∘)=−tan‌18∘, we have:
tan‌66∘×tan‌6∘=‌

tan‌18∘

tan‌54∘

Multiplying equations (ii) and (iii):
tan‌6∘‌tan‌42∘‌tan‌66∘‌tan‌78∘=‌

tan‌54∘

tan‌18∘

×‌

tan‌18∘

tan‌54∘

=1
Therefore, tan‌6∘+tan‌42∘+tan‌66∘+tan‌78∘ sums up to a neat result based on symmetrical properties of angles and tangent function identities.