TS EAMCET 10 May 2024 Shift 1 Solved Paper

Given:
cos284∘+sin‌2126∘−sin‌84∘‌cos‌126∘=K
We can simplify this as follows:
‌sin‌2126∘=sin‌2(90∘+36∘)=cos236∘‌ by using the identity ‌sin‌(90∘+θ)=cos‌θ
‌−sin‌84∘‌cos‌126∘=cos(90∘−84∘)‌cos(90∘+36∘)
Substituting these identities, we get:
K=cos284∘+cos236∘−sin‌84∘‌cos‌126∘
Next, using the identity for product-to-sum, sin‌θ‌cos‌φ=‌

1

2

[sin‌(θ+φ)+sin‌(θ−φ)], substitute values as needed, ultimately simplifying:
K=cos284∘−sin‌236∘+1+‌

3

4

−sin‌24∘
Continuing to simplify using trigonometric identities:
K=cos‌120∘‌cos‌48∘+‌

7

4

−sin‌224∘
Now solve for K :
K=−‌

1

2

(1−2sin‌224∘)+‌

7

4

−sin‌224∘
This results in:
K=−‌

1

2

+‌

7

4

=‌

5

4

Now, for the expression involving tan‌A :
tan‌A+cot‌A=2K

Substitute K=‌

5

4

:
tan‌A+‌

1

tan‌A

=‌

5

2

Multiply through by tan‌A to eliminate the fraction:
2tan2A−5‌tan‌A+2=0
Factor the quadratic equation:
(2‌tan‌A−1)(tan‌A−2)=0
Thus, the possible values for tan‌A are:
tan‌A=‌

1

2

,2