Subject

    Bank Exam

    Topic

    Joint Recruitment Test for 7 Banks and 1 Financial Institutions - (Senior Officers/General) - 27.11.2021

    If 1+sinθ=xcosθ,then tanθ=?

    ক)
    (x²+1)/x
    খ)
    (x²-1)/x
    গ)
    (x²+1)/2x
    ঘ)
    (x²-1)/2x

    Explanation

    দেওয়া আছে , 1+sinθ=xcosθ ⇒ (1+sinθ)/cosθ=x ⇒ 1/cosθ+sinθ/cosθ=x ∴ secθ+tanθ=x..............i আমরা জানি , sec²θ-tan²θ=1 ⇒ (secθ+tanθ)(secθ-tanθ)=1 ⇒ x(secθ-tanθ)=1 ⇒ secθ-tanθ=1/x ⇒ secθ-1/x=tanθ ⇒ x-tanθ-1/x=tanθ ⇒ x-1/x=tanθ+tanθ ⇒ (x²-1)/2x=tanθ ∴ tanθ=(x²-1)/2x

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