Question 2 tan A -1- A -tan A -1- A -2 cosec A

LHS=tan A/1+sec A tan A/1 sec A=tan A(1 sec A 1 sec A)/(1+sec A)(1 sec A) =tan A( 2 sec A)/(1 sec^2 A) = 2 tan A. sec A/(sec^2 A 1) [∵(a+b)(a b)=a^2 b^2] ...

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Standard X

Mathematics

Trigonometric Ratios

Question 2 ta...

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Question 2
$\frac{t a n A}{1 + s e c A} - \frac{t a n A}{1 - s e c A} = 2 c o s e c A$

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Solution

$L H S = \frac{t a n A}{1 + s e c A} - \frac{t a n A}{1 - s e c A} = \frac{t a n A (1 - s e c A - 1 - s e c A)}{(1 + s e c A) (1 - s e c A)}$

$= \frac{t a n A (- 2 s e c A)}{(1 - s e c^{2} A)} = \frac{2 t a n A . s e c A}{(s e c^{2} A - 1)} [∵ (a + b) (a - b) = a^{2} - b^{2}]$

$= \frac{2 t a n A . s e c A}{t a n^{2} A} [∵ s e c^{2} A - t a n^{2} A = 1]$

$[∵ s e c θ = \frac{1}{c o s θ} a n d t a n θ = \frac{s i n θ}{c o s θ}]$

$= \frac{2 s e c A}{t a n A} = \frac{2}{s i n A} = 2 c o s e c A = R H S [∵ c o s e c θ = \frac{1}{s i n θ}]$

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Question 2 tan A1+sec A−tan A1−sec A=2 cosec A

Question 2
$\frac{t a n A}{1 + s e c A} - \frac{t a n A}{1 - s e c A} = 2 c o s e c A$