Question

can you plse help me in this question


tanθ/1-cotθ+cotθ/1-tanθ=1+tanθ+cotθ​

Answer 1

Answer:

the answer will be

Step-by-step explanation:

\tan( \alpha ) (1 - \tan( \alpha ) ) + \cot( \alpha )(1 + \cot( \alpha )tan(α)(1−tan(α))+cot(α)(1+cot(α)

tan-tan²+cot-cot²/1(1-tan) -cot(1-tan)

tan+cot-(tan²+cot²)/1-tan-cot+cottan

tan+cot-1/1-tan-cottan just solve this u get urs answer perfectly

Answer 2

$</p><p></p><p>LHS = tanθ/(1 - cotθ) + cotθ/(1 - tanθ) </p><p></p><p></p><p>= tanθ/(1 - 1/tanθ) + (1/tanθ)/(1 - tanθ)</p><p></p><p></p><p>= tan²θ/(tanθ - 1) + 1/tanθ(1 - tanθ)</p><p></p><p></p><p>= tan³θ/(tanθ - 1) - 1/tanθ(tanθ - 1) </p><p></p><p></p><p>= (tan³θ - 1)/tanθ(tanθ - 1) </p><p></p><p></p><p>= (tanθ - 1)(tan²θ + 1 + tanθ)/tanθ(tanθ - 1) </p><p></p><p></p><p>= (tan²θ + 1 + tanθ)/tanθ</p><p> </p><p></p><p>= tanθ + cotθ + 1 </p><p></p><p></p><p>= sinθ/cosθ + cosθ/sinθ + 1 </p><p></p><p></p><p>= (sin²θ + cos²θ)/sinθ.cosθ + 1 </p><p></p><p></p><p>= secθ.cosecθ + 1 </p><p></p><p></p><p>= 1 + secθ.cosecθ = RHS</p><p></p><p>$