Question

1+cosθ+sinθ/1+cosθ-sinθ = 1+sinθ/cosθ

Answer 1

Step-by-step explanation:

LHS

1+cosA+sinA/1+cosA-sinA×1+cosA+sinA/1+cosA+sinA

={(1+cosA)+sinA}²/(1-cosA)²-sin²A

=(1+cosA)²+sin²A+2(1+cosA)sinA/1+2cosA+cos²A-sin²A

=1+cos²A+2cosA+sin²A+2sinA+2sinA.cosA/1+2cosA+cos²A-1+cos²A

=2+2cosA+2sinA+2sinAcosA/2cosA+2cos²A

=2(1+cosA+sinA+sinAcosA)/2(cosA+cos²A)

=(1+cosA)+sinA(1+cosA/cosA(1+cosA)

=(1+cosA)(1+sinA)/cosA(1+cosA

=1+sinA/cosA

RHS

Answer 2

Answer:

✶⊶⊷⊶⊷❍ ❥ ❍⊶⊷⊶⊷✶

$\large\green{\mid{\fbox{\tt{Ꭲօ թɾօѵҽ}}\mid}}$

$\frac{sinθ}{1+cosθ}+\frac{1+cosθ}{sinθ}=2cosecθ$

$\large\red{\mid{\fbox{\tt{รοℓυƭเօɳ}}\mid}}$

$\large\pink{\mid{\fbox{\tt{ᏞᎻร}}\mid}}$=

$⟹\sf\bold{\blue{\frac{sinθ}{1+cosθ}+\frac{1+cosθ}{sinθ}}}$

$⟹\sf\bold{\blue{\frac{sin²+(1+cosθ)²}{(1+cosθ)sinθ}}}$

$⟹\sf\bold{\blue{\frac{sin²+1+cos²θ+2cosθ}{(1+cosθ)sinθ}}}$

$⟹\sf\bold{\blue{\frac{sin²+cos²θ+1+2cosθ}{(1+cosθ)sinθ}}}$

$⟹\sf\bold{\blue{\frac{2+2cosθ}{(1+cosθ)sinθ}}}$

$⟹\sf\bold{\blue{\frac{2(1+cosθ)}{(1+cosθ)sinθ}}}$

$⟹\sf\bold{\blue{\frac{2}{sinθ}}}$

$⟹\large\pink{\mid{\fbox{\tt{2.coseθ}}\mid}}$

$\large\pink{\mid{\fbox{\tt{ᏞᎻร=ƦᎻร}}\mid}}$

✶⊶⊷⊶⊷❍ ❥ ❍⊶⊷⊶⊷✶

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⠀⠀⠀ $\large\green{\mid{\fbox{\tt{❥ϐℓυєᴇყεร}}\mid}}$

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