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Posted by Mansi R 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
We have,
{tex} \mathrm { LHS } = \frac { \sec \theta - \tan \theta } { \sec \theta + \tan \theta }{/tex}
{tex} \Rightarrow \quad \mathrm { LHS } = \frac { \sec \theta - \tan \theta } { \sec \theta + \tan \theta } \times \frac { \sec \theta - \tan \theta } { \sec \theta - \tan \theta }{/tex}
{tex} \Rightarrow \quad \mathrm { LHS } = \frac { ( \sec \theta - \tan \theta ) ^ { 2 } } { \sec ^ { 2 } \theta - \tan ^ { 2 } \theta } = \frac { ( \sec \theta - \tan \theta ) ^ { 2 } } { 1 }{/tex} [{tex} \because{/tex} sec2{tex} \theta{/tex} - tan2{tex} \theta{/tex} = 1]
{tex} \Rightarrow \quad \mathrm { LHS } = \sec ^ { 2 } \theta - 2 \sec \theta \tan \theta + \tan ^ { 2 } \theta{/tex}
{tex} \Rightarrow \quad \mathrm { LHS } = \left( 1 + \tan ^ { 2 } \theta \right) - 2 \sec \theta \tan \theta + \tan ^ { 2 } \theta{/tex} [{tex} \because{/tex} sec2{tex} \theta{/tex} = 1 + tan2{tex} \theta{/tex}]
{tex} \Rightarrow{/tex} LHS = 1 - 2sec{tex} \theta{/tex} tan{tex} \theta{/tex} + 2tan2{tex} \theta{/tex} = RHS
Posted by Abhishek Gupta 6 years, 9 months ago
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Posted by Siksha Upadhyay 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let coordinate of P be (0, y) and of Q be (x,0)
A(2, -5) is mid - point of PQ.
By Section Formula,
(2, -5) = {tex}\left( \frac { 0 + x } { 2 } , \frac { y + 0 } { 2 } \right){/tex}
{tex}\therefore 2=\frac x2 \text{ and }-5=\frac y2{/tex}
x = 4 and y = -10.
Therefore, P is (0, -10) and Q is (4, 0).
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Noor Mehta 6 years, 9 months ago
0Thank You