Question

Maintenance of machine in a factory can be carried out by one man or sometimes by a two man crew.

The times taken with one man crew are 1.5, 2.0, 2.5 or 3.0 hours with probabilities OF 0.20, 0.30, 0.35,

and 0.15 43wp3ctively. A two man crew requires 0.75, 1.00, 1.50 and 2.00 hours with the probabilities of

0.25, 0.35, 020, and 0.20 respectively. The cost of labour is Rs. 7 per hour and the overhead expenses are

Rs. 3 per hour per crew. Decide whether one or two man crew is more economical.​

Answer 1

$\huge\colorbox{pink}{Meditation}$

Meditation is a practice where an individual uses a technique – such as mindfulness, or focusing the mind on a particular object, thought, or activity – to train attention and awareness, and achieve a mentally clear and emotionally calm and stable state.

Answer 2

Step-by-step explanation:

★

To prove:−

\sf \frac{Sin\theta -cos\theta +1}{sin\theta +cos\theta -1}=\frac{1}{sec\theta - tan\theta}

sinθ+cosθ−1

Sinθ−cosθ+1

=

secθ−tanθ

1

\begin{gathered}\\\end{gathered}

\begin{gathered}\bigstar \underline{\underline{\sf Proof:-}}\\\end{gathered}

★

Proof:−

\begin{gathered}\\\end{gathered}

Formulas used:-

\begin{gathered}\\\end{gathered}

\begin{gathered}\leadsto \sf \frac{sin\theta }{cos\theta } =tan\theta\\\end{gathered}

⇝

cosθ

sinθ

=tanθ

\begin{gathered}\leadsto \sf \frac{1}{cos\theta } =sec\theta \\\end{gathered}

⇝

cosθ

1

=secθ

\begin{gathered}\leadsto \sf sec^2\theta -tan^2\theta =1\\\end{gathered}

⇝sec

2

θ−tan

2

θ=1

\begin{gathered}\leadsto \sf a^2 -b^2 =(a+b)(a-b)\\\end{gathered}

⇝a

2

−b

2

=(a+b)(a−b)

-----------------------

LHS:

\begin{gathered}:\implies \sf \frac{ \sf sin\theta - cos\theta +1}{sin\theta +cos\theta -1}\\\\\end{gathered}

:⟹

sinθ+cosθ−1

sinθ−cosθ+1

Dividing numerator & denominator by cosθ:

\begin{gathered}:\implies \sf \frac{(sin\theta - cos\theta +1)/cos\theta}{(sin\theta +cos\theta -1)/cos\theta} \\\\\end{gathered}

:⟹

(sinθ+cosθ−1)/cosθ

(sinθ−cosθ+1)/cosθ

\begin{gathered}:\implies \sf \frac{(sin\theta /cos\theta-(cos\theta /cos\theta )+1/cos\theta}{(sin\theta /cos\theta ) +(cos\theta /cos\theta )-1/cos\theta } \\\\\end{gathered}

:⟹

(sinθ/cosθ)+(cosθ/cosθ)−1/cosθ

(sinθ/cosθ−(cosθ/cosθ)+1/cosθ

\begin{gathered}:\implies \sf \frac{tan\theta -1 +sec\theta }{tan\theta +1-sec\theta } \\\\\end{gathered}

:⟹

tanθ+1−secθ

tanθ−1+secθ

\begin{gathered}:\implies \sf \frac{tan\theta +sec\theta -1}{tan\theta -sec\theta +(sec^2\theta - tan^2\theta )} \\\\\end{gathered}

:⟹

tanθ−secθ+(sec

2

θ−tan

2

θ)

tanθ+secθ−1

\begin{gathered}:\implies \sf \frac{tan\theta +sec\theta -1}{(tan\theta -sec\theta )+(sec\theta -tan\theta )(sec\theta +tan\theta )} \\\\\end{gathered}

:⟹

(tanθ−secθ)+(secθ−tanθ)(secθ+tanθ)

tanθ+secθ−1

\begin{gathered}:\implies \sf \frac{\cancel{tan\theta +sec\theta -1}}{sec\theta -tan\theta \cancel{-1+sec\theta +tan\theta }} \\\\\end{gathered}

:⟹

secθ−tanθ

−1+secθ+tanθ

tanθ+secθ−1

\begin{gathered}:\implies \bold{\sf \frac{1}{sec\theta tan\theta }}\\\\\end{gathered}

:⟹

secθtanθ

1

Therefore,

\begin{gathered}\mapsto \boxed{\boxed{\sf \frac{sin\theta - cos\theta + 1}{sin\theta + cos\theta - 1}= \frac{1}{sec\theta - tan\theta } }}\\\\\end{gathered}

↦

sinθ+cosθ−1

sinθ−cosθ+1

=

secθ−tanθ

1

Hence proved.

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HOPE IT HELPS!