Question

prove that tanθ/1-Cotθ + cotθ/1-tanθ = 1 + secθ . cosθ.​

Answer 1

plzzzzzzzzzzz...... mark as brainliest

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Correction in Question on RHS there should be 1 + secθ cosecθ

LHS :-

$\tt{\rightarrow\dfrac{tan\theta}{1-cot\theta}+\dfrac{cot\theta}{1-tan\theta}}$

$\tt{\rightarrow\dfrac{sin\theta/cos\theta}{1-cos\theta/sin\theta}+\dfrac{cos\theta/sin\theta}{1-sin\theta/cos\theta}}$

$\tt{\rightarrow\dfrac{sin^2\theta}{cos\theta(sin\theta-cos\theta)}+\dfrac{cos^2\theta}{sin\theta(cos\theta-sin\theta)}}$

$\tt{\rightarrow\dfrac{sin^3\theta-cos^3\theta}{cos\theta\times sin\theta(sin\theta-cos\theta)}}$

${\boxed{\sf\:{Identity=a^3-b^3=(a-b)(a^2+ab+b^2)}}}$

$\tt{\rightarrow\dfrac{(sin\theta-cos\theta)(sin^2\theta+sin\theta cos\theta+cos^2\theta)}{cos\theta sin\theta(sin\theta-cos\theta)}}$

${\boxed{\sf\:{Using=sin^2\theta+cos^2\theta=1}}}$

$\tt{\rightarrow\dfrac{1+sin\theta cos\theta}{cos\theta sin\theta}}$

$\tt{\rightarrow\dfrac{1}{cos\theta sin\theta}+1}$

= secθ cosecθ + 1

= 1 + secθ cosecθ

LHS = RHS