Question

prove sinA (1+tanA) +cosA(1+cotA) =sinA+cosec A​

Answer 1

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To prove:

sin A (1+tan A) +cos A (1+cot A) =sec A+cosec A

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Left Hand side:

sinA (1+tanA) +cosA (1+cotA)

 sinA + sinA tanA + cosA + cosA tanA

 sinA + sinA $\frac{sinA}{cosA }$ + cosA + $\frac{cosA}{sinA}$ cosA {∵  $tan A=\frac{sinA}{cosA }$ & $cot A = \frac{cosA}{sinA}$ }

 sinA + $\frac{sin^{2}A }{cosA}$ + cosA + $\frac{cos^{2} A}{sinA}$

 $\frac{sin^{2} AcosA+cos^{2} AsinA+sin^{3}A+cos^{3}A}{sinA cosA}$

{Multiplying with numerator}

 $\frac{sinAcosA(sinA+cosA)sin^{3}A cos^{3}A }{sinAcosA}$

{ Taking sin and cos common}

 $\frac{sinAcosA(sinA+cosA)(sin^{2} A+cos^{2} A-cosAsinA) }{sinAcosA}$

{∵ a³+b³ = (a+b)(a²+b²-ab)}

 $\frac{sinAcosA(sinA+cosA)(1-cosAsinA)}{sinAcosA }$  

{ ∵ Sin²A + Cos²A = 1}

  $\frac{sinA+ cosA.1 }{sinAcosA}$

{ sinAcosA got cancelled}

 $\frac{sinA}{sinAcosA} + \frac{cosA}{sinAcosA }$

 $\frac{1}{cosA} + \frac{1}{sinA }$

= secA + cosecA {R.H.S}

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

This formulas will be we used in 11th standard trigonometry.Opposite angle can be calculated by using graphs I,II,III and IV th quadrant can be related as

I = A = All are positive

II = S = Sin and Cosec are positive

III = T = Tan and Cot are positive

IV = C = Cos and Sec are positive

Hope it helps

Thanks for asking