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प्रश्न
Prove that `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec(90^circ - A) cosec(90^circ - A)`
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उत्तर
LHS = `sinA/sin(90^circ - A) + cosA/cos(90^circ - A)`
⇒ `cos(90^circ - A)/sin(90^circ - A) + sin(90^circ - A)/cos(90^circ - A)`
⇒ `(cos^2(90^circ - A) + sin^2(90^circ - A))/(sin(90^circ - A) . cos(90^circ - A)) = 1/(sin(90^circ - A) . cos(90^circ - A))`
⇒ `sec(90^circ - A) cosec(90^circ - A)`
APPEARS IN
संबंधित प्रश्न
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