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Question Number 1206 by gs8763330@gmail.com last updated on 14/Jul/15

prove that 2sin^(−1) 3/4=tan^(−1) 24/7

$${prove}\:{that}\:\:\mathrm{2sin}^{−\mathrm{1}} \mathrm{3}/\mathrm{4}=\mathrm{tan}^{−\mathrm{1}} \mathrm{24}/\mathrm{7} \\ $$

Commented by 123456 last updated on 14/Jul/15

sin α=a⇔^(b.c) α=sin^(−1) a cos α=b⇔^(b.c) α=cos^(−1) b tan α=((sin α)/(cos α))=t⇔^(b.c) α=tan^(−1) ((sin α)/(cos α))=tan^(−1) (a/b)=tan^(−1) t

$$\mathrm{sin}\:\alpha={a}\overset{{b}.{c}} {\Leftrightarrow}\alpha=\mathrm{sin}^{−\mathrm{1}} {a} \\ $$$$\mathrm{cos}\:\alpha={b}\overset{{b}.{c}} {\Leftrightarrow}\alpha=\mathrm{cos}^{−\mathrm{1}} {b} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\alpha}={t}\overset{{b}.{c}} {\Leftrightarrow}\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\alpha}=\mathrm{tan}^{−\mathrm{1}} \frac{{a}}{{b}}=\mathrm{tan}^{−\mathrm{1}} {t} \\ $$

Answered by prakash jain last updated on 14/Jul/15

sin^(−1) (3/5)=θ sin θ=(3/5), cos θ=(4/5) tan 2θ=((2sin θcos θ)/(cos^2 θ−sin^2 θ))=((2∙(3/5)∙(4/5))/(((16)/(25))−(9/(25))))=((24)/7) ⇒2θ=tan^(−1) ((24)/7) 2sin^(−1) (3/5)=tan^(−1) ((24)/7) There is a typo in your question. It should be 2sin^(−1) (3/5)=tan^(−1) ((24)/7)

$$\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{5}}=\theta \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{3}}{\mathrm{5}},\:\mathrm{cos}\:\theta=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{tan}\:\mathrm{2}\theta=\frac{\mathrm{2sin}\:\theta\mathrm{cos}\:\theta}{\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{sin}^{\mathrm{2}} \theta}=\frac{\mathrm{2}\centerdot\frac{\mathrm{3}}{\mathrm{5}}\centerdot\frac{\mathrm{4}}{\mathrm{5}}}{\frac{\mathrm{16}}{\mathrm{25}}−\frac{\mathrm{9}}{\mathrm{25}}}=\frac{\mathrm{24}}{\mathrm{7}} \\ $$$$\Rightarrow\mathrm{2}\theta=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{24}}{\mathrm{7}} \\ $$$$\mathrm{2sin}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{5}}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{24}}{\mathrm{7}} \\ $$$$\mathrm{There}\:\mathrm{is}\:\mathrm{a}\:\mathrm{typo}\:\mathrm{in}\:\mathrm{your}\:\mathrm{question}.\:\mathrm{It} \\ $$$$\mathrm{should}\:\mathrm{be} \\ $$$$\mathrm{2sin}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{5}}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{24}}{\mathrm{7}} \\ $$

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