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If sin theta + cos theta = m and sec th...

If ` sin theta + cos theta = m and sec theta + "cosec" theta = n, ` prove that
` n(m^(2)-1) = 2m. `

Text Solution

Verified by Experts

We have
` n(m^(2) -1) = ( sec theta + "cosec" theta )[( sin theta + cos theta)^(2) -1] `
` = ((1)/(cos theta) + (1)/(sin theta))[(sin^(2)theta + cos^(2)theta) + 2 sin theta cos theta -1 ] `
` =((sin theta + cos theta))/(sin theta cos theta) * 2 sin theta cos theta `
`=2(sin theta + cos theta)= 2m. `
Hence, `n(m^(2) -1) = 2m. `
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Knowledge Check

  • If sin theta + costheta = m and sin theta - cos theta = n , then

    A
    `m^(2) + n^(2) = 1`
    B
    `m^(2) - n^(2) = 1`
    C
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  • If sin theta + cos theta = m and theta + cosec theta = n , then the value of n(m^(2) - 1) is

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    m
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    2m
    C
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    D
    2 n

  • If sin theta + cosec theta =2, then sin^n theta+"cosec"^n theta is equal to

    A
    2
    B
    2n
    C
    `2^(n-1)`
    D
    `2^(n-2)`

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