Prove -1 - cos A- - sin A - -sin A- - -1 - cos A- - 2 cosec A- - Maths Q - A

Proof:Consider,L.H.S=(1+cos A) /sin A+(sin A)/(1+cos A)0ex0ex = (1+cos A)(1+cos A)+(sin A)(sin A) /sin A(1+cos A)0ex0ex = (1+cos A)^2+ (sin A ...

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Projection Formula

Prove (1 + co...

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Prove $\frac{(1 + \cos A)}{\sin A} + \frac{(\sin A)}{(1 + \cos A)} = 2 cosec A$

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ω

\frac{1}{a + ω} + \frac{1}{b + ω} + \frac{1}{c + ω} = 2 ω^{2}

\frac{1}{a + ω^{2}} + \frac{1}{b + ω^{2}} + \frac{1}{c + ω^{2}} = 2 ω

\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1}

\frac{x}{a} + \frac{y}{b} = 1

\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c^{2}}

c

P_{1}, P_{2}, P_{3}

A, B, C

Δ

\frac{1}{P_{1}} + \frac{1}{P_{2}} - \frac{1}{P_{3}} = \frac{2 a b}{(a + b + c) Δ} c o s^{2} \frac{C}{2}

I f

A

(A - 2 I) (A - 4 I) = 0

\frac{1}{6} A + \frac{4}{3} A^{- 1} =

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Δ

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