Question

cotA+tanA=cosecA . secA , is the statement is true or false​

Answer 1

Question :

cotA+tanA=cosecA . secA

Formula used :

• sin²A + cos²A = 1
• sec²A - tan²A = 1
• cosec²A - cot²A = 1

Solution :

LHS

=cotA+tanA

$= \frac{ \cos(A) }{ \sin(A) } + \frac{ \sin(A) }{ \cos(A) }$

$= \frac{ \cos {}^{2} (A) + \sin {}^{2} (A) }{ \sin(A) \cos(A) }$

$= \frac{1}{ \sin(A) \cos(A) }$

= cosecA . secA

RHS =cosecA . secA

⇒LHS = RHS

$\huge{\bold{ Hence\: Proved}}$

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More Trigonometry Formulas:

1. sin2A = 2 sinA cosA
2. cos2A = cos²A - sin²A
3. tan2A = 2 tanA / (1 - tan²A)

Answer 2

❥${\purple{\underline{\underline{\large{\mathtt{QUESTION:-}}}}}}$

Prove that,

$\sf{cotA+tanA\:=\: cosecA.\:secA}$

❥${\purple{\underline{\underline{\large{\mathtt{SOLUTION:-}}}}}}$

Taking L.H.S,

$\sf{\:\:\:cotA+tanA}$

★We know $\sf{cotA=\frac{cosA}{sinA}}$and $\sf{tanA=\frac{sinA}{cosA}}$★

$\sf\implies{\frac{cosA}{sinA}+\frac{sinA}{cosA}}$

$\sf\implies{\frac{cos^2+sin^2}{sinA.\:cosA}}$

★We know sin²A + cos²A = 1★

$\sf\implies{\frac{1}{sinA.\:cosA}}$

$\sf\implies{\frac{1}{sinA}\:.\:\frac{1}{cosA}}$

★We know $\sf{\frac{1}{sinA}=cosecA}$ and $\sf{\frac{1}{cosA}=secA}$★

$\sf\implies{cosecA\:.\:secA}$

L.H.S = R.H.S ( Proved )

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Some formulas related to trigonometry :-

•$\sf{1+tan^2A=sec^2A}$

•$\sf{1+cot^2A=cosec^2A}$

•$\sf{sinA=\sqrt{1-cos^2A}}$

•$\sf{cosA=\sqrt{1-sin^2A}}$

•$\sf{secA=\sqrt{1+tan^2A}}$

•$\sf{tanA=\sqrt{sec^2A-1}}$

•$\sf{cosecA=\sqrt{1+cot^2A}}$

•$\sf{cotA=\sqrt{cosec^2A-1}}$

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