UPSC CDS 1 2022 Math Paper

Given:
1. √sec2θ+cosec2θ=tan‌θ+cot‌θ
2. √tan2θ+cot2θ+4=sec‌θ+cosecθ
Concept used:
sec2θ−tan2θ=1
cosec2θ−cot2θ=1
cosecθ=1∕sin‌θ
(a+b)2=a2+b2+2ab
Calculation:
Statement 1:
√sec2θ+cosec2θ=tan‌θ+cot‌θ
⇒sec2θ+cosec2θ=(tan‌θ+cot‌θ)2
We know that, (a+b)2=a2+2ab+b2
⇒sec2θ+cosec2θ=tan2θ+cot2θ+2‌tan‌θ⋅cot‌θ
⇒sec2θ−tan2θ+cosec2θ−cot2θ=2‌tan‌θ×1∕cot‌θ
⇒1+1=2
⇒2=2‌ [Satisying] ‌
Statement 2:
√tan2θ+cot2θ+4=sec‌θ+cosecθ
⇒tan2θ+cot2θ+4=(sec‌θ+cosecθ)2
⇒tan2θ+cot2θ+4=sec2θ+cosec2θ+2‌sec‌θ×cosecθ
⇒tan2θ−sec2θ+cot2θ−cosec2θ+4=2‌sec‌θ×cosecθ
⇒−1−1+4=2‌sec‌θ×cosecθ
⇒2=2‌sec‌θ×cosecθ
⇒1=sec‌θ×cosecθ
⇒sin‌θ‌cos‌θ=1 which is not possible because 0<θ<90∘
∴ Only 1 is the correct statement.