Question

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Answer 1

✿ 34 Part One

$\mathsf{Given : (sinA - cosA)(1 + cotA + tanA)}$

$\implies \textsf{sinA + sinA.cotA + sinA.tanA - cosA - cosA.cotA - cosA.tanA}$

$\bullet\;\;\textsf{Rearrange terms in such a way which throws light on our solution}$

$\implies \textsf{(sinA - cosA) + (sinA.cotA - cosA.tanA) + (sinA.tanA - cosA.cotA)}$

$\bullet\;\;\textsf{We can notice that our required answer is the last part.}\\\\\bullet\;\;\textsf{So, We need to eliminate the first two parts}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{cot\theta = \dfrac{cos\theta}{sin\theta}}}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{tan\theta = \dfrac{sin\theta}{cos\theta}}}}$

$\bullet\;\;\textsf{Substitute values of cotA and tanA only in the middle part}$

$\implies \mathsf{(sinA - cosA) + \bigg(sinA.\dfrac{cosA}{sinA} - cosA.\dfrac{sinA}{cosA}\bigg) + (sinA.tanA - cosA.cotA)}$

$\implies \mathsf{(sinA - cosA) + (cosA - sinA) + (sinA.tanA - cosA.cotA)}$

$\implies \mathsf{(sinA.tanA - cosA.cotA)}$

✿ 34 Part Two

★  sec(90 - θ) = cosecθ

$\mathsf{\implies sec29^{\circ} = sec(90^{\circ} - 61^{\circ}) = cosec61^{\circ}}$

$\mathsf{\implies \dfrac{sec29^{\circ}}{cosec61^{\circ}} = \dfrac{cosec61^{\circ}}{cosec61^{\circ}} = 1}$

★  cot(90 - θ) = tanθ

$\mathsf{\implies cot8^{\circ} = cot(90^{\circ} - 82^{\circ}) = tan82^{\circ}}$

$\mathsf{\implies cot17^{\circ} = cot(90^{\circ} - 73^{\circ}) = tan73^{\circ}}$

★  sin(90 - θ) = cosθ

$\mathsf{\implies sin52^{\circ} = sin(90^{\circ} - 38^{\circ}) = cos38^{\circ}}$

$\mathsf{\implies sin^252^{\circ} = cos^238^{\circ}}$

★  tanθ.cotθ = 1

★  sin²θ + cos²θ = 1

★  cot45° = 1

The Question changes to :

$\mathsf{\implies 1 + 2tan82^{\circ}.tan73^{\circ}.cot73^{\circ}.cot82^{\circ} - 3(sin^238^{\circ} + cos^238^{\circ})}$

$\mathsf{\implies 1 + 2tan82^{\circ}.cot82^{\circ}.tan73^{\circ}.cot73^{\circ} - 3}$

$\mathsf{\implies 1 + 2 - 3}$

$\mathsf{\implies 3 - 3}$

$\mathsf{\implies 0}$