UPSC CDS 1 2022 Math Paper

Given:
tan‌θ+sec‌θ=3
Formula Used:
sec2θ−tan2θ=1
(sec‌θ+tan‌θ)(sec‌θ−tan‌θ)=1
Calculation:
We have tan‌θ+sec‌θ=3‌‌⋯ (i)
According to the formula used
(sec‌θ+tan‌θ)(sec‌θ−tan‌θ)=1
⇒3(sec‌θ−tan‌θ)=1
⇒(sec‌θ−tan‌θ)=1∕3‌‌‌ - (ii) ‌
On adding (i) and (ii), we get
2‌sec‌θ=3+1∕3=10∕3
⇒sec‌θ=5∕3‌‌⋯‌-(iii) ‌
On putting the value of sec‌θ in (i), we get
tan‌θ+5∕3=3
⇒tan‌θ=3−5∕3=4∕3‌‌‌ - -(iv) ‌
Now, we have to find the value of 3‌tan‌θ+9‌sec‌θ
So, from (iii) and (iv), we get
3×4∕3+9×5∕3
⇒4+15=19
∴ The required value of 3‌tan‌θ+9‌sec‌θ is 19 .