Question

$$\begin{gathered} Proovethat \\ 37.\sin ({270}^0+A)=-\cos A,and\tan ({270}^0+A)=-cotA \end{gathered}$$

Answer 1

Dear Student,
$$\begin{gathered} 37. \\ \sin \left(270°+A\right)=-\cos A \\ L.H.S \\ \sin \left(270°+A\right)=\sin \left[180°+90°+A\right] \\ =\sin \left[180°+\left(90°+A\right)\right] \\ =-\sin \left(90°+A\right)[\sin ce\sin (180°+\theta )=-\sin \theta ] \\ therefore,\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\left(\mathbf{270}\mathbf{°}\mathbf{+}\mathbf{A}\right)\mathbf{=}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathit{A}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}[sincesin(90°+\theta )=cos\theta ] \\ \tan (270°+A)=\tan [180°+90°+A] \\ =\tan [180°+(90°+A\left)\right] \\ =\tan (90°+A)[\sin ce\tan (180°+\theta )=\tan \theta ] \\ Therefore,\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathbf{(}\mathbf{270}\mathbf{°}\mathbf{}\mathbf{+}\mathbf{}\mathit{A}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{}\mathit{c}\mathit{o}\mathit{t}\mathbf{}\mathit{A}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}[\sin ce\tan (90°+\theta )=-cot\theta ] \end{gathered}$$

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