Question

plz help. me to solve this​

Answer 1

★To prove:-

• $\displaystyle \sf \frac{tan\theta }{1-cot\theta } +\frac{cot \theta }{1-tan\theta } = 1+sec\theta .cosec\theta$

★Proof:-

LHS:

$\implies \displaystyle \sf \frac{tan\theta }{1-cot\theta } +\frac{cot\theta }{1-tan\theta }$

Putting[✦cotθ = 1/tanθ]

$\\ :\implies \displaystyle \sf \frac{tan\theta }{1-\frac{1}{tan\theta} } + \frac{\frac{1}{tan\theta } }{1-tan\theta}$

$\displaystyle \implies \sf \frac{tan\theta }{\dfrac{tan\theta - 1}{tan\theta }} + \frac{1}{tan\theta (1-tan\theta )}$

$\displaystyle \implies \sf \frac{tan^2\theta }{tan\theta - 1} +\frac{1}{tan\theta(1-tan\theta )}$

$\displaystyle \implies \sf \frac{tan^2\theta }{tan\theta -1} \bold{-} \frac{1}{tan\theta (tan\theta -1)}$

Taking 1/(tanθ-1) common,

$\displaystyle \implies \sf \frac{1}{(tan\theta -1)} \bigg[tan^2\theta - \frac{1}{tan\theta } \bigg]$

$\displaystyle \implies \sf \frac{1}{(tan\theta -1)} \bigg[ \frac{tan^3\theta -1}{tan\theta } \bigg]$

Using the formula,

✦a³ - b³ = (a-b)(a²+b²+ab)

$\\ \displaystyle \implies \sf \frac{1}{\cancel{(tan\theta -1)}} \bigg[\frac{\cancel{(tan\theta -1)}(tan^2\theta +1+tan\theta )}{tan\theta }\bigg]$

$\displaystyle \implies \sf \frac{tan^2\theta +1+tan\theta }{tan\theta }$

$\displaystyle \implies \sf \frac{tan^2\theta }{tan\theta } +\frac{1}{tan\theta } +\frac{tan\theta }{tan\theta }$

$\displaystyle \implies \sf tan\theta +cot\theta + 1$

Using the formula's,

✦tanθ = sinθ/cosθ

✦cotθ = cosθ/sinθ

$\\ \displaystyle \implies \sf \frac{sin\theta }{cos\theta } + \frac{cos\theta }{sin\theta } +1$

$\displaystyle \implies \sf \frac{sin^2\theta +cos^2\theta }{sin\theta cos\theta }+1$

As [✦sin²θ + cos²θ = 1]

$\displaystyle \implies \sf \frac{1}{sin\theta cos\theta } +1$

Also,

✦1/sinθ = cosecθ

✦1/cosθ = secθ

$\\ \displaystyle \implies \sf \bigg( \frac{1}{sin\theta } \times \frac{1}{cos\theta }\bigg) +1$

$\displaystyle \implies \sf \underline{\boxed{\bold{1+sec\theta .cosec\theta }}}$

      = RHS

Hence proved!

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Answer 2

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Hope It Helps!