Prove that-A-1- A-1-sin A-1-cos A2- A-cosec A2

L.H.S. = sec A 1/sec A+1 = 1/cos A 1/1/cos A+1 = 1 cos A/cos A/1+cos A/cos A = 1 cos A/1+cos A Multiply and divide by (1 + cos A) = (1 cos A)(1+cos A)/(1+cos ...

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Standard X

Mathematics

Trigonometric Identity- 1

Prove that:A-...

Question

Prove that:
$\frac{s e c A - 1}{s e c A + 1} = (\frac{s i n A}{1 + c o s A})^{2} = (c o t A - c o s e c A)^{2}$

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Solution

$L . H . S . = \frac{s e c A - 1}{s e c A + 1} = \frac{\frac{1}{c o s A} - 1}{\frac{1}{c o s A} + 1} = \frac{\frac{1 - c o s A}{c o s A}}{\frac{1 + c o s A}{c o s A}} = \frac{1 - c o s A}{1 + c o s A}$
Multiply and divide by (1 + cos A)
= $\frac{(1 - c o s A) (1 + c o s A)}{(1 + c o s A) (1 + c o s A)} = \frac{1 - c o s^{2} A}{(1 - c o s A)^{2}} = \frac{s i n^{2} A}{(1 + c o s A)^{2}}$
$= (\frac{s i n A}{1 + c o s A})^{2}$
And, $(\frac{s i n A}{1 + c o s A})^{2} = [(\frac{s i n A}{1 + c o s A}) \times \frac{1 - c o s A}{1 - c o s A}]^{2}$
$= (\frac{(s i n A) (1 - c o s A)}{1 - c o s^{2} A})^{2}$
$= (\frac{(s i n A) (1 - c o s A)}{s i n^{2} A})^{2}$
$= (\frac{(1 - c o s A)}{s i n A})^{2}$
$= (c o s e c A - c o t A)^{2}$
$= (c o t A - c o s e c A)^{2}$
Hence Proved

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Prove that: sec A−1sec A+1=(sin A1+cos A)2=(cot A−cosec A)2

Prove that:
$\frac{s e c A - 1}{s e c A + 1} = (\frac{s i n A}{1 + c o s A})^{2} = (c o t A - c o s e c A)^{2}$