Question

please help class 10 maths​

Answer 1

To prove:-

$\rm\dagger\quad \dfrac{cot \theta}{1+tan\theta} = \dfrac{cot \theta -1}{2-sec^2 \theta}$

Solution

$\rm\dashrightarrow\quad \dfrac{cot \theta}{1+tan\theta} = \dfrac{cot \theta -1}{2-sec^2 \theta}$

Multiply 1 - tan θ on numerator and denominator in LHS

$\rm\dashrightarrow \dfrac{cot \theta(1 - tan \theta)}{(1+tan\theta)(1-tan\theta)} = \dfrac{cot \theta -1}{2-sec^2 \theta}$

We know that,

cot θ = 1 ÷ tan θ

Hence, cot θ tan θ = 1

$\rm\dashrightarrow \dfrac{cot \theta - 1}{1-tan^2\theta} = \dfrac{cot \theta -1}{2-\sec^2 \theta} \qquad \qquad ...\bigg [ \because (a+b)(a-b)=a^2-b^2 \bigg ]$

We know that,

sec² θ - tan²θ = 1

Hence, tan² θ = sec² θ - 1

$\rm\dashrightarrow \dfrac{cot \theta - 1}{1-(\sec^2\theta -1)} = \dfrac{cot \theta -1}{2-sec^2 \theta}$

$\rm\dashrightarrow \dfrac{cot \theta - 1}{1-sec^2\theta +1} = \dfrac{cot \theta -1}{2-sec^2 \theta}$

$\rm\dashrightarrow \dfrac{cot \theta - 1}{2-sec^2\theta} = \dfrac{cot \theta -1}{2-sec^2 \theta}$

Hence, Proved.