If sec-m and tan-n- then 1-m-m-n-1-m-n- is equals to

Explanation for the correct option Substitute the value of sec(θ)=m and tan(θ)=n in 1/m[(m+n)+1/(m+n)], then[ 1/m[(m+n)+1/(m+n)]=1/sec(θ)[sec(θ)+tan(θ)+1/sec(θ) ...

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Standard XIII

Mathematics

Quadratic Equation

If secθ=m and...

Question

If $\sec (θ) = m$ and $\tan (θ) = n$ , then $\frac{1}{m} [(m + n) + \frac{1}{(m + n)}]$ is equals to

$2$

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$2 m$

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$2 n$

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$m n$

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Solution

The correct option is A
$2$

Explanation for the correct option
Substitute the value of $\sec (θ) = m$ and $\tan (θ) = n$ in $\frac{1}{m} [(m + n) + \frac{1}{(m + n)}]$ , then
$\begin{matrix} \frac{1}{m} [(m + n) + \frac{1}{(m + n)}] = \frac{1}{\sec (θ)} [\sec (θ) + \tan (θ) + \frac{1}{\sec (θ) + \tan (θ)}] \\ = \frac{1}{\sec (θ)} [\frac{\sec^{2} (θ) + \tan^{2} (θ) + 2 \tan (θ) \sec (θ) + 1}{\sec (θ) + \tan (θ)}] \end{matrix}$
Put trigonometry identity $t a n^{2} (θ) = s e c^{2} (θ) - 1$
$\begin{array}{rcl} \frac{1}{m} [(m + n) + \frac{1}{(m + n)}] & = & \frac{1}{\sec (θ)} [\frac{\sec^{2} (θ) + \sec^{2} (θ) - 1 + 2 \tan (θ) \sec (θ) + 1}{\sec (θ) + \tan (θ)}] \\ = & \frac{2 \sec^{2} (θ) + 2 \sec (θ) \tan (θ)}{\sec (θ) [\sec (θ) + \tan (θ)]} \end{array}$
Take common $2 \sec (θ)$ in the numerator,
$\begin{array}{rcl} \frac{1}{m} [(m + n) + \frac{1}{(m + n)}] & = & \frac{2 \sec (θ) [\sec (θ) + \tan (θ)]}{\sec (θ) [\sec (θ) + \tan (θ)]} \\ = & 2 \end{array}$
The value of $\frac{1}{m} [(m + n) + \frac{1}{(m + n)}]$ is equal to $2$
Hence, option $(A)$ is correct.

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If secθ=m and tanθ=n, then 1mm+n+1m+n is equals to

If $\sec (θ) = m$ and $\tan (θ) = n$ , then $\frac{1}{m} [(m + n) + \frac{1}{(m + n)}]$ is equals to