Question

Ch- trigonometry pls no spam​

Answer 1

EXPLANATION.

⇒ cos(A).cot(A)/1 - sin(A) = 1 + cosec(A).

As we know that,

From L.H.S we get,

⇒ cos(A).cos(A)/sin(A)/1 - sin(A).

⇒ cos²(A)/sin(A)/1 - sin(A).

⇒ cos²(A)/sin(A)(1 - sin(A)).

⇒ (1 - sin²A)/(sin(A))(1 - sin(A)).

⇒ (1 - sin(A)(1 + sin(A)/sin(A)(1 - sin(A)).

⇒ (1 + sin(A))/sin(A).

⇒ [1/sin(A) + sin(A)/sin(A)].

⇒ cosec(A) + 1.

Hence Proved.

                                                                                                                     

MORE INFORMATION.

(1) = sin²θ + cos²θ = 1.

(2) = 1 + tan²θ = sec²θ.

(3) = 1 + cot²θ = cosec²θ.

Answer 2

$\sf{\bold{\blue{\underline{\underline{Given}}}}}$

• $\sf{\dfrac{cosA.cotA}{1-sinA}=1+cosecA   }$  ⠀⠀⠀⠀

$\sf{\bold{\red{\underline{\underline{To\:Find}}}}}$

• LHS=RHS⠀⠀⠀⠀

$\sf{\bold{\purple{\underline{\underline{Solution}}}}}$

LHS:-

:$\rightsquigarrow$ $\sf{\dfrac{cosA.cotA}{1-sinA}}$

:$\implies$ $\bold{\dfrac{cosA \dfrac{cosA}{sinA}}{1-sinA}}$ 

:$\hookrightarrow$ $\tt{\dfrac{\dfrac{cos^{2}A}{sinA}}{1-sinA   }}$ 

:$\longrightarrow$ $\sf{\dfrac{cos^{2}A}{sinA(1-sinA)}   }$ 

:$\dashrightarrow$ $\bold{\dfrac{ 1-sin^{2}A}{sinA(1-sinA)}  }$ 

:$\rightsquigarrow$ $\tt{\dfrac{(1+sinA)(1-sinA)}{sinA(1-sinA)}  }$

:$\implies$ $\sf{\dfrac{(1+sinA)\cancel{(1-sinA)}}{sinA\cancel{(1-sinA)}}  }$

:$\longrightarrow$ $\bold{\dfrac{1+sinA}{sinA}   }$ 

:$\dashrightarrow$ $\tt{\dfrac{1}{sinA}+\dfrac{sinA}{sinA}   }$ 

:$\maltese$$\sf{\blue{cosecA+1}   }$=RHS

$\therefore{\red{LHS=RHS }}$(proved)