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Prove that :
[tex]\sf{\dfrac{tan\theta+sec\theta-1}{tan\theta-sec\theta+1}= sec\theta+tan\theta=\dfrac{1+sin\theta}{cos\thet

By Autumn

Prove that :
[tex]\sf{\dfrac{tan\theta+sec\theta-1}{tan\theta-sec\theta+1}= sec\theta+tan\theta=\dfrac{1+sin\theta}{cos\theta}[/tex]

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5. Using the formula, calculate the amount, when:
Principal = 34,000, Rate = 17% per annum, Time = 2 years.​

Find the value of k If the quadratic
equation. Kx(X-4)+16=0 has equal roots .​

1 thought on “Prove that :<br /> [tex]\sf{\dfrac{tan\theta+sec\theta-1}{tan\theta-sec\theta+1}= sec\theta+tan\theta=\dfrac{1+sin\theta}{cos\thet”

  1. Answer:

    Step-by-step explanation:

    To Prove :-

    [tex]\dfrac{tan\theta+sec\theta-1}{tan\theta-sec\theta+1}=sec\theta+tan\theta=\dfrac{1+sin\theta}{cos\theta}[/tex]

    Solution :-

    Taking L.H.S :-

    [tex]= \dfrac{tan\theta+sec\theta-1}{tan\theta-sec\theta+1}[/tex]

    [tex]=\dfrac{tan\theta+sec\theta-(sec^2\theta-tan^2\theta)}{tan\theta-sec\theta+1}[/tex]

    [tex]=\dfrac{tan\theta+sec\theta-((sec\theta+tan\theta)(sec\theta-tan\theta))}{tan\theta-sec\theta+1}[/tex]

    [tex]=\dfrac{sec\theta+tan\theta(1-(sec\theta-tan\theta))}{tan\theta-sec\theta+1}[/tex]

    [tex]=\dfrac{sec\theta+tan\theta(1-sec\theta+tan\theta)}{tan\theta-sec\theta+1}[/tex]

    [tex]=sec\theta+tan\theta[/tex]

    [tex]=\dfrac{1}{cos\theta}+\dfrac{sin\theta}{cos\theta}[/tex]

    [tex]=\dfrac{1+sin\theta}{cos\theta}[/tex]

    = R.H.S

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