Question

Prove that ;
(1+cotA-cosecA)* (1+tanA+secA)=2​

Answer 1

S O L U T I O N :

Prove that;

(1+cot A - cosec A) × (1+tan A + sec A) = 2

$\underline{\bf{Explanation\::}}$

Firstly, as we know that formula & some changes;

• cot Ф = cos Ф / sin Ф
• cosec Ф = 1 / sin Ф
• tan Ф = sin Ф / cos Ф
• sec Ф = 1 / cos Ф

Now;

Taking L.H.S :

$\mapsto\tt{(1+cot\:A - cosec\:A) (1+tan\:A+sec\:A) }$

$\mapsto\tt{\bigg(1+\dfrac{cos\:A}{sin\:A} - \dfrac{1}{sin\:A} \bigg) \bigg(1+\dfrac{sin\:A}{cos\:A} +\dfrac{1}{cos\:A}\bigg)}$

$\mapsto\tt{\dfrac{1}{sin\:A} \bigg(sin\:A+cos\:A - 1\bigg) \dfrac{1}{cos\:A} \bigg(cos\:A + sin\:A + 1\bigg)\:\:\underbrace{\sf{taking\:common\:as\:1/sin\:A \:\:\&\:\:1/cos\:A}}}$$\mapsto\tt{\dfrac{1}{sin\:Acos\:A} \bigg[(sin\:A+cos\:A)^{2} -(1)^{2}\bigg]}$

$\mapsto\tt{\dfrac{1}{sin\:Acos\:A} \bigg[(sin\:A)^{2} +(cos\:A)^{2} + 2\times sin\:A\times cos\:A-1\bigg]\:\:\underbrace{\sf{Using\:formula \:of\:(a+b)^{2}}}}$$\mapsto\tt{\dfrac{1}{sin\:Acos\:A} \bigg[\cancel{1} + 2sin\:A cos\:A-\cancel{1} \bigg]\:\:\underbrace{\sf{\therefore \:sin^2A + cos^2 A = 1}}}$

$\mapsto\tt{\dfrac{1}{sin\:Acos\:A} \bigg[ 2sin\:A cos\:A\bigg]}$

$\mapsto\tt{\dfrac{2\cancel{\:sin\:A cos\:A}}{\cancel{sin\:Acos\:A} }}$

$\mapsto\bf{2}$

Thus;

The L.H.S = R.H.S .

It's proved .