Question

1÷tanA - 1÷tan2A = cosec2A​

Answer 1

Step-by-step explanation:

1

−

tan2A

1

=csc2A

L.H.S. = \dfrac{1}{\tan A } -\dfrac{1}{\tan 2A }

tanA

1

−

tan2A

1

Using the trigonometric identity,

\tan 2A = \dfrac{2\tan A}{1-\tan^2 A}tan2A=

1−tan

2

A

2tanA

= \dfrac{1}{\tan A } -\dfrac{1}{\dfrac{2\tan A}{1-\tan^2 A}}

tanA

1

−

1−tan

2

A

2tanA

1

=\dfrac{1}{\tan A } -\dfrac{1-\tan^2 A}{2\tan A}=

tanA

1

−

2tanA

1−tan

2

A

Taking LCM of denominator part, we get

=\dfrac{2-1+\tan^2 A}{2\tan A}=

2tanA

2−1+tan

2

A

=\dfrac{1+\tan^2 A}{2\tan A}=

2tanA

1+tan

2

A

=\dfrac{1}{\dfrac{2\tan A}{1+\tan^2 A} }=

1+tan

2

A

2tanA

1

Using the trigonometric identity,

\sin 2A=\dfrac{2\tan A}{1+\tan^2 A}sin2A=

1+tan

2

A

2tanA

=\dfrac{1}{\sin 2A}=

sin2A

1

= \csc 2Acsc2A

= R.H.S., proved.

∴ \dfrac{1}{\tan A } -\dfrac{1}{\tan 2A } =\csc 2A

tanA

1

−

tan2A

1

=csc2A , proved

Answer 2

$\large\sf\underline{Question}$

• $\sf\:\frac{1}{tan(A)}-\frac{1}{tan(2A)}= Cosec(2A)$

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‎⠀⠀━━━━━ ★ ━━━━━

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$\large\sf\underline{Solution}$

$\sf➪\:\frac{1}{tan(A)}-\frac{1}{tan(2A)}$

$\tt\red{tan(2A)=\frac{2tan(A)}{1+tan^{2}(A)}}$

$\sf➪\:\frac{1}{tan(A)}-\frac{1 \div 2 tan (A) }{1+tan^{2}(A)}$

$\sf➪\:\frac{1}{tan(A)}-\frac{1+tan^{2}(A)}{2tan(A)}$

$\sf➪\:\frac{2-1+tan^{2}(A)}{2tan(A)}$

$\sf➪\:\frac{1+tan^{2}(A)}{2tan(A)}$

$\tt\red{sin(2A)=\frac{2tan(A)}{1+tan^{2}(A)}}$

$\sf➪\:\frac{1}{sin(2A)}$

$\tt\red{Cosec(A)=\frac{1}{sin(A)}}$

$\sf➪\:Cosec(2A)\:[proved]$

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‎⠀⠀━━━━━ ★ ━━━━━

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$\fbox\pink{S}\fbox\green{O}\fbox\pink{M}\fbox\green{E}\fbox\pink{F}\fbox\green{O}\fbox\pink{R}\fbox\green{M}\fbox\pink{U}\fbox\green{L}\fbox\pink{A}\red{࿐}$

• $\sf\:sin^{2}θ+cos^{2}θ=1$

• $\sf\:cosec^{2}θ-cot^{2}θ=1$

• $\sf\:sec^{2}θ-tan^{2}θ=1$

• $\sf\:cos2θ=2cos^{2}θ-1$

• $\sf\:sin2θ=2sinθ cosθ$

Thnku ❣