Math, asked by TheMahakals, 10 months ago

Answer it ................

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Answered by Anonymous

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Answered by BRAINLYADDICTOR

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$<marquee behaviour-move><font color="yellow black"><h1>ANSWER:</ ht></marquee>$

$(s = z) \\ \tanθ \div 2 = \sqrt{1 - e \: \div 1 + e } \tanθ \div 2 \\ \tanθ \div 2 = \sqrt{1 - e \:\div 1 + e}tanθ \div 2 \\ s.o.b.s \\ \tan {}^{2}θ \div 2 =$

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