Question

( cosecx - cotx )^2 = 1 - cosx / 1 + cosx​

Answer 1

Answer:

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Answer 2

Solution :

$\bf{(cosec\:x - cot\:x)^{2} = \dfrac{1 - cos\:x}{1 + cos\:x}}$

From the above equation , we get that :

• LHS = $\bf{(cosec\:x - cot\:x)^{2}}$

• RHS = $\bf{\dfrac{1 - cos\:x}{1 + cos\:x}}$

By solving the LHS , we get :

$:\implies \bf{(cosec\:x - cot\:x)^{2}} \\ \\ \\ :\implies \bf{cosec^{2}x - 2 \times cosec\:x \times cot\:x + cot^{2}x} \\ \\ \bf{\because\:\:(a + b)^{2} = a^{2} + 2ab + b^{2}} \\ \\ \\ :\implies \bf{cosec^{2}x - 2cosec\:x cot\:x + cot^{2}x} \\ \\ \\ :\implies \bf{\bigg(\dfrac{1}{sin\:x}\bigg)^{2} - 2 \times \dfrac{1}{sin\:x} \times \dfrac{cos\:x}{sin\:x} + \bigg(\dfrac{cos\:x}{sin\:x}\bigg)^{2}}$

$:\implies \bf{\dfrac{1}{sin^{2}x} - \dfrac{2cos\:x}{sin^{2}x} + \dfrac{cos^{2}x}{sin^{2}x}} \\ \\ \\ :\implies \bf{\dfrac{1 - 2cos\:x + cos^{2}x}{sin^{2}x}} \\ \\ \\ :\implies \bf{\dfrac{1 - (cos\:x + cos\:x) + cos^{2}x}{sin^{2}x}} \\ \\ \\ :\implies \bf{\dfrac{1 - cos\:x - cos\:x + cos^{2}x}{sin^{2}x}}$

$:\implies \bf{\dfrac{1(1 - cos\:x) - cos\:x(1 - cos\:x)}{sin^{2}x}} \\ \\ \\ :\implies \bf{\dfrac{(1 - cos\:x)(1 - cos\:x)}{sin^{2}x}} \\ \\ \\ :\implies \bf{\dfrac{(1 - cos\:x)(1 - cos\:x)}{1 - cos^{2}x}\quad [sin^{2}\theta = 1 - cos^{2}\theta} \\ \\ \\ :\implies \bf{\dfrac{(1 - cos\:x)(1 - cos\:x)}{1^{2} - cos^{2}x}} \\ \\ \bf{\because\:\:(a - b)^{2} = (a + b)(a - b)}$

$:\implies \bf{\dfrac{(1 - cos\:x)(1 - cos\:x)}{(1 + cos\:x)(1 - cos\:x)}} \\ \\ \\ :\implies \bf{\dfrac{1 - cos\:x}{1 + cos\:x}} \\ \\ \\ \boxed{\therefore \bf{LHS = \dfrac{1 - cos\:x}{1 + cos\:x}}}$

Hence the LHS of the equation is $\bf{\dfrac{1 - cos\:x}{1 + cos\:x}}$

By putting the LHS and RHS together , we get :

$:\implies \bf{LHS = RHS} \\ \\ \\ :\implies \bf{\dfrac{1 - cos\:x}{1 + cos\:x} = \dfrac{1 - cos\:x}{1 + cos\:x}} \\ \\ \\ \boxed{\therefore \bf{(cosec\:x - cot\:x)^{2} = \dfrac{1 - cos\:x}{1 + cos\:x}}}$

Proved !!