Question

1-cosA/1+cosA=(cotA-cosecA)2^​

Answer 1

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Prove:

$\dfrac{1-cosA}{1+ cosA } = (Cot A- Cosec A)^{2}$

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Left Hand Side:

$= \dfrac{1-cosA}{1+ cosA }$  

Take the Conjugate of 1 + cos A {i.e 1 - cos A}

$= \dfrac{1-cosA}{1+ cosA } \times \dfrac{1-cosA}{1- cosA }$

$= \dfrac{(1-cosA)^{2} }{1-cos^{2}A}$

Use the identity 1 - cos²A = sin²A

$=\dfrac{(1-cosA)^{2}}{sin^{2}A }$

$=[\dfrac {1-cosA}{sinA}]^{2}$

$= (\dfrac{1}{sinA} - \dfrac{cosA}{sinA})^{2}$

$= (CosecA - CotA )^{2}$

$\pink{ \boxed{Hence Proved }}$

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To solve all trigonometric equation we need to know all trigonometric identities and its functions

They are as follows,

✪ Reciprocal Identities ✪

✧ Sin θ = 1/Cosec θ

✧ Cos θ = 1/Sec θ

✧ Tan θ =  1/cot θ

✪ Pythagorean Identities ✪

✧ Sin²θ + Cos²θ = 1

✧ 1 + tan²θ = sec²θ

✧ 1 + cot²θ = cosec²θ

✪ Ratio Identities ✪

✧ Tan θ = Sin θ/Cos θ

✧ Cot θ = Cos θ/Sin θ

✪ Complementary Angles Identities ✪

✧ Sin (90 – θ) = Cos θ

✧ Cos (90 – θ) = Sin θ

✧ Tan (90 – θ) = Cot θ

✧ Cot ( 90 – θ) = Tan θ

✧ Sec (90 – θ) = Cosec θ

✧ Cosec (90 – θ) = Sec θ

✪ Opposite Angle Identities ✪

✧ Sin (-θ) = – Sin θ

✧ Cos (-θ) = Cos θ

✧ Tan (-θ) = – Tan θ

✧ Cot (-θ) = – Cot θ

✧ Sec (-θ) = Sec θ

✧ Cosec (-θ) = -Cosec θ

ℌope ℑt ℌelps

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