Question

cot A-1/2-sec^2 A=cot A/1+tan A
Prove this. ​

Answer 1

Answer:

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Answer 2

Answer:

$\Large{\underline{\underline{\bf{Given:}}}}$

$\dfrac{cot\:A-1}{2-sec^{2}\:A } =\dfrac{cot\:A}{1+tan\:A}$

$\Large{\underline{\underline{\bf{To\:Prove:}}}}$

• LHS = RHS

$\Large{\underline{\underline{\bf{Identities\:used:}}}}$

→ (a + b) × (a - b) = a² - b²

→ cot A × tan A = 1

→ tan² A = sec² A - 1

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Consider the RHS

  $\dfrac{cot\:A}{1+tan\:A}$

→ Taking the conjugate of 1 + tan A and multiplying on both numerator and denominator.

  $\dfrac{cot\:A}{1+tan\:A} \times \dfrac{1-tan\:A}{1-tan\:A}$

$=\dfrac{cot\:A-cot\:A\:tan\:A}{1-tan^{2}\:A }$

$=\dfrac{cot\:A-1}{1-(sec^{2}\:A-1) }$

$=\dfrac{cot\:A-1}{1+1-sec^{2}\:A }$

$=\dfrac{cot\:A-1}{2-sec^{2} \:A}$

= LHS

→ Hence proved.

$\Large{\underline{\underline{\bf{More\:Identities:}}}}$

→ sin² A + cos² A = 1

→ cosec² A - cos² A = 1

→ sin A × cosec A = 1

→ cos A × sec A = 1