Math, asked by arjunkcsuresh974, 3 months ago

Prove that tan 45 - tan 3a/1+tan45×tan 30 =(2- 3​

Answers

Answered by Xmart34
1

Step-by-step explanation:

x²-3abx+2b²=0

or, {(ax)²-2.ax.3b/2+(3b/2)²}+2b²-9b²/4=0

or, (ax-3b/2)²+(8b²-9b²)/4=0

or, (ax-3b/2)²-(b/2)²=0

or, (ax-3b/2+b/2)(ax-3b/2-b/2)=0

or, (ax-b)(ax-2b)=0

Either, ax-b=0

or, ax=b

Answer:

SOLUTION :–

• Let the function –

\begin{gathered} \\ \implies\bf y = { \tan}^{ - 1} (x) \\ \end{gathered}

⟹y=tan

−1

(x)

• Let the function –

\begin{gathered} \\ \implies\bf x= \tan( \theta) \\ \end{gathered}

⟹x=tan(θ)

• Differentiate with respect to 'θ' –

\begin{gathered} \\\implies\bf \dfrac{dx}{d \theta}= \sec^{2} ( \theta) \\ \end{gathered}

dx

=sec

2

(θ)

\begin{gathered} \\\implies\bf \dfrac{dx}{d \theta}= 1 + \tan^{2} ( \theta) \\ \end{gathered}

dx

=1+tan

2

(θ)

\begin{gathered} \\\implies\bf \dfrac{dx}{d \theta}= 1 +x^{2} \: \: \: \: \: - - - eq.(1)\\ \end{gathered}

dx

=1+x

2

−−−eq.(1)

• Now –

\begin{gathered} \\ \implies\bf y = { \tan}^{ - 1} ( \tan( \theta) ) \\ \end{gathered}

⟹y=tan

−1

(tan(θ))

\begin{gathered} \\ \implies\bf y = \theta\\ \end{gathered}

⟹y=θ

• Differentiate with respect to 'θ' –

\begin{gathered} \\ \implies\bf \dfrac{dy}{d \theta} = \dfrac{d\theta}{d\theta}\\ \end{gathered}

dy

=

\begin{gathered} \\ \implies\bf \dfrac{dy}{d \theta} =1\\ \end{gathered}

dy

=1

• We should write this as –

\begin{gathered} \\ \implies\bf \dfrac{dy}{dx} \times \dfrac{dx}{d \theta} =1\\ \end{gathered}

dx

dy

×

dx

=1

• Using eq.(1) –

\begin{gathered} \\ \implies\bf \dfrac{dy}{dx} (1 + {x}^{2})=1\\ \end{gathered}

dx

dy

(1+x

2

)=1

\begin{gathered} \\ \large \implies{ \boxed{\bf \dfrac{dy}{dx} = \dfrac{1}{(1+{x}^{2})}}}\\ \end{gathered}

dx

dy

=

(1+x

2

)

1

or, x=b/a

Or, ax-2b=0

Answer:

SOLUTION :–

• Let the function –

\begin{gathered} \\ \implies \bf y = log_{ex}{logx} \\ \end{gathered}

⟹y=log

ex

logx

• Using identity –

\begin{gathered} \\ \implies \bf log_{a}{b} = \dfrac{ log_{e}(b) }{ log_{e}(a) } \\ \end{gathered}

⟹log

a

b=

log

e

(a)

log

e

(b)

• So –

\begin{gathered} \\ \implies \bf y = \dfrac{log_{e}(logx)}{ log_{e}(ex)}\\ \end{gathered}

⟹y=

log

e

(ex)

log

e

(logx)

• Using identity –

\begin{gathered} \\ \implies \bf log_{e}(a.b) = log_{e}(a)+log_{e}(b) \\ \end{gathered}

⟹log

e

(a.b)=log

e

(a)+log

e

(b)

• So –

\begin{gathered} \\ \implies \bf y = \dfrac{log_{e}(logx)}{ log_{e}(e) +log_{e}(x)}\\ \end{gathered}

⟹y=

log

e

(e)+log

e

(x)

log

e

(logx)

\begin{gathered} \\ \implies \bf y = \dfrac{log_{e}(logx)}{ 1 +log_{e}(x)}\\ \end{gathered}

⟹y=

1+log

e

(x)

log

e

(logx)

• Now Differentiate with respect to 'x' –

\begin{gathered} \\ \implies \bf \dfrac{dy}{dx} = \dfrac{[1 +log_{e}(x)] \bigg \{\dfrac{1}{x log_{e}(x)} \bigg \}- [log_{e}(logx)] \bigg( \dfrac{1}{x} \bigg) }{ [1 +log_{e}(x)]^{2} }\\ \end{gathered}

dx

dy

=

[1+log

e

(x)]

2

[1+log

e

(x)]{

xlog

e

(x)

1

}−[log

e

(logx)](

x

1

)

\begin{gathered} \\ \implies \bf \dfrac{dy}{dx} = \dfrac{\bigg \{\dfrac{1}{x log_{e}(x)}+ \dfrac{1}{x} \bigg\}- \dfrac{log_{e}(logx)}{x}}{ [1 +log_{e}(x)]^{2} }\\ \end{gathered}

dx

dy

=

[1+log

e

(x)]

2

{

xlog

e

(x)

1

+

x

1

}−

x

log

e

(logx)

\begin{gathered} \\ \implies \bf \dfrac{dy}{dx} = \dfrac{\dfrac{1}{x} \bigg \{\dfrac{1}{ log_{e}(x)}+1 - log_{e}(logx)\bigg\}}{ [1 +log_{e}(x)]^{2} }\\ \end{gathered}

dx

dy

=

[1+log

e

(x)]

2

x

1

{

log

e

(x)

1

+1−log

e

(logx)}

or, ax=2b

or, x=2b/a

∴, the roots of the given equations are: b/a, 2b/a. Ans.

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s helps u࿐

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