Question

Proof that cosec theta minus cot theta whole square equal to 1 minus cos theta divided by 1 + cos theta

Answer 1

Correct Question :-

$\rm\:{(cosec\theta-cot\theta)}{^2}=\dfrac{1-cos\theta}{1+cos\theta}$

Solution :-

$\rm\:LHS={(cosec\theta-cot\theta)}^{2}$

$\rm\longrightarrow\:\bigg(\dfrac{1}{sin\theta}-\dfrac{cos\theta}{sin\theta}\bigg)^2$

$\rm\longrightarrow\:\bigg(\dfrac{1-cos\theta}{sin\theta}\bigg)^2$

$\rm\longrightarrow\:\dfrac{(1-cos\theta)^2}{(si{n}^{2}\theta)}$

$\rm\longrightarrow\:\dfrac{(1-cos\theta)\:(1-cos\theta)}{(1-co{s}^{2}\theta)}$

$\rm\longrightarrow\:\dfrac{(1-cos\theta)(1-cos\theta)}{{(1)}^{2}-{(cos\theta)^2}}$

$\rm\longrightarrow\:\dfrac{(\cancel{1-cos\theta})(1-cos\theta)}{(\cancel{1-cos\theta})(1+cos\theta)}$

$\rm\longrightarrow\:\dfrac{1-cos\theta}{1+cos\theta}$

$\rm\:LHS=RHS$

$\rm\:Hence,proved$

❍ Some Trigonometry formula :-

• sinθ . cosecθ = 1

• cosθ . secθ = 1

• tanθ . cotθ = 1

• sin²θ + cos²θ = 1

• sec²θ - tan²θ = 1

• cosec² - cot² = 1