App for CBSE class 10, Karnataka SSLC, and BSc Nursing

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Specific angles & Trigonometric ratios	$$\theta =0°$$	$$\theta =30°$$	$$\theta =45°$$	$$\theta =60°$$	$$\theta =90°$$
$$sin\theta$$	$$0$$	$$\frac{1}{2}$$	$$\frac{1}{\sqrt{2}}$$	$$\frac{\sqrt{3}}{2}$$	$$1$$
$$cos\theta$$	$$1$$	$$\frac{\sqrt{3}}{2}$$	$$\frac{1}{\sqrt{2}}$$	$$\frac{1}{2}$$	$$0$$
$$tan\theta$$	$$0$$	$$\frac{1}{\sqrt{3}}$$	$$1$$	$$\sqrt{3}$$	$$Notdefined$$
$$cosec\theta$$	$$Notdefined$$	$$2$$	$$\sqrt{2}$$	$$\frac{2}{\sqrt{3}}$$	$$1$$
$$sec\theta$$	$$1$$	$$\frac{2}{\sqrt{3}}$$	$$\sqrt{2}$$	$$2$$	$$Notdefined$$
$$cot\theta$$	$$Notdefined$$	$$\sqrt{3}$$	$$1$$	$$\frac{1}{\sqrt{3}}$$	$$0$$

$$sin\theta =cos(90°-\theta )$$	$$cos\theta =sin(90°-\theta )$$	$$tan\theta =cot(90°-\theta )$$
$$cosec\theta =sec(90°-\theta )$$	$$sec\theta =cosec(90°-\theta )$$	$$cot\theta =tan(90°-\theta )$$

$$si{n}^2\theta +co{s}^2\theta =1$$

$$1+ta{n}^2\theta =se{c}^2\theta$$

$$1+co{t}^2\theta =cose{c}^2\theta$$

$$co{s}^2\theta =1-si{n}^2\theta$$	$$se{c}^2\theta -ta{n}^2\theta =1$$	$$cose{c}^2\theta -co{t}^2\theta =1$$
$$si{n}^2\theta =1-co{s}^2\theta$$	$$se{c}^2\theta -1=ta{n}^2\theta$$	$$cose{c}^2\theta -1=co{t}^2\theta$$

$sin\theta$ in terms of $cos\theta ,tan\theta$	$$sin\theta =\sqrt{1-co{s}^2\theta }$$	$$sin\theta =\frac{tan\theta }{\sqrt{1-ta{n}^2\theta }}$$
$cos\theta$ in terms of $sin\theta ,tan\theta$	$$cos\theta =\sqrt{1-si{n}^2\theta }$$	$$cos\theta =\frac{1}{\sqrt{1+ta{n}^2\theta }}$$
$tan\theta$ in terms of $sin\theta ,cos\theta$	$$tan\theta =\frac{sin\theta }{\sqrt{1-si{n}^2\theta }}$$	$$cos\theta =\frac{\sqrt{1+co{s}^2\theta }}{cos\theta }$$