Question

solve the above questions⬆️​

Answer 1

Answer:

(1)$\frac{1}{\sqrt{2} }$

(2)$\sqrt{2}$

(3)1

Step-by-step explanation:

From the figure, AB = BC = a and AC = $\sqrt{2} a$ .[Pythagorus theorem]

(1) sin A =$\frac{opposite}{hypotenuse}=\frac{a}{\sqrt{2}a } =\frac{1}{\sqrt{2} }$

(2)sec A = $\frac{hypotenuse}{adjacent}= \frac{\sqrt{2}a }{a}= \sqrt{2}$

(3)$sin^{2}x+cos^{2}x =1$ [Trigonometric identity]

Answer 2

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