If A and B are complementary angles, prove that : (i) cot B + cos B = sec A cos B (1 + sin B) (ii) cot A cot B - sin A cos B - cos A sin B = 0 (iii) cosec² A + cosec² B = cosec² A cosec² B (iv)

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KnowledgeBoat · Accepted Answer

Given, A + B = 90° B = 90° - A and A = 90° - B. (i) Substituting value of B in L.H.S. of equation : By formula, sin (90° - B) = cos B and cos (90° - B) = sin B ⇒ cos B sec A (1 + sin B). Since, L.H.S. = R.H.S. Hence, proved that cot B + cos B = sec A cos B (1 + sin B). (ii) Substituting value of B in L.H.S. of equation : ⇒ cot A cot (90° - A) - sin A cos (90° - A) - cos A sin (90° - A) By formula, cot (90° - A) = tan A, cos (90° - A) = sin A and sin (90° - A) = cos A ⇒ cot A tan A - sin A sin A - cos A cos A ⇒ cot A - (sin² A + cos² A) ⇒ 1 - 1 ⇒ 0. Since, L.H.S. = R.H.S. Hence, proved that cot A cot B - sin A cos B - cos A sin B = 0. (iii) Substituting value of B in L.H.S. of equation : ⇒ cosec² A + cosec² (90° - A) By formula, cosec (90° - A) = sec A ⇒ + sec² A ⇒ ⇒ By formula, sin² A + cos² A = 1 and sec (90° - A) = cosec A ⇒ ⇒ cosec² A sec² A ⇒ cosec² A sec² (90° - B) ⇒ cosec² A cosec² B Since, L.H.S. = R.H.S. Hence, proved that cosec² A + cosec² B = cosec² A cosec² B. (iv) Solving L.H.S. of the equation : By formula, cos (90° - θ) = sin θ and sin (90° - θ) = cos θ. By formula, cos² A = 1 - sin² A Since, L.H.S. = R.H.S. Hence, proved that .

Mathematics

If A and B are complementary angles, prove that : (i) cot B + cos B = sec A cos B (1 + sin B) (ii) cot A cot B - sin A cos B - cos A sin B = 0 (iii) cosec<sup>2</sup> A + cosec<sup>2</sup> B = cosec<sup>2</sup> A cosec<sup>2</sup> B (iv)

Trigonometric Identities

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