Question

If given that p1 and p2 are the lengths of the perpendiculars drawn from the points (cosA , sinA) and (-secA, cosecA) on the line x secA+ y cosecA = 0, respectively, prove that (4/p²1)-P2² = 4.​

Answer 1

Step-by-step explanation:

We have,

$\color{cyan}{ \sf{p_{1} = \dfrac{ \left| cos(A) \cdot sec(A) + sin(A) \cdot \: cosec(A)\right| }{sec^{2} (A) + cosec^{2}(A) } }}$

$\color{blue}{ \implies \sf{p_{1} = \dfrac{ \left| 1 + 1\right| }{ \dfrac{1}{cos^{2} (A)} + \dfrac{1}{sin^{2}(A)} } }}$

$\color{blue}{ \implies \sf{p_{1} = \dfrac{ \left| 1 + 1\right| }{ \dfrac{ sin^{2} (A) + cos^{2} (A) }{cos^{2} (A) \cdot \: sin^{2}(A)} } }}$

$\color{blue}{ \implies \sf{p_{1} = \dfrac{ 2 \: sin^{2} (A) \: cos^{2}(A) }{ sin^{2} (A) + cos^{2} (A) } }}$

$\color{blue}{ \implies \sf{p_{1} = 2 \: sin^{2} (A) \: cos^{2}(A) }}$

$\color{blue}{ \implies \sf{p_{1} = \dfrac{4\: sin^{2} (A) \: cos^{2}(A) }{2}}}$

$\color{blue}{ \implies \sf{p_{1} = \dfrac{sin^{2} (2A) }{2}}}$

$\color{blue}{ \implies \sf{2 \: p_{1} = sin^{2} (2A) \: \: \: \: \: \: \: \: \: ...(2) }}$

$\color{orange}{ \sf{p_{2} = \dfrac{ \left| - sec(A) \cdot sec(A) + cosec(A) \cdot \: cosec(A)\right| }{sec^{2} (A) + cosec^{2}(A) } }}$

$\color{blue}{ \implies \sf{p_{2} = \dfrac{ \left| - sec^{2} (A) + cosec^{2} (A) \right| }{sec^{2} (A) + cosec^{2}(A) } }}$

$\color{blue}{ \implies \sf{p_{2} = \dfrac{ \left| - \dfrac{1}{cos^{2} (A) } + \dfrac{1}{sin^{2} (A)} \right| }{\dfrac{1}{cos^{2} (A) } + \dfrac{1}{sin^{2} (A)}} }}$

$\color{blue}{ \implies \sf{p_{2} = \dfrac{ \left| \dfrac{cos^{2} (A) - sin^{2} (A) }{cos^{2} (A) \cdot \: sin^{2} (A) } \right| }{\dfrac{cos^{2} (A) + sin^{2} (A) }{cos^{2} (A) \cdot \: sin^{2} (A) }} }}$

$\color{blue}{ \implies \sf{p_{2} = \dfrac{ \left| cos^{2} (A) - sin^{2} (A) \right| }{cos^{2} (A) + sin^{2} (A) } }}$

$\color{blue}{ \implies \sf{p_{2} = cos (2A) }}$

$\color{blue}{ \implies \sf{{p_{2}} ^{2} = cos^{2} (2A) \: \: \: \: \: \: \: \: \: \: ...(2)}}$

Adding (1) and (2),

$\rm{2 \: p_{1} +{ p_{2} }^{2} = 1}$