Question

pls help.......fastt plss​

Answer 1

$\huge\tt{\red{Given:}}$

★$(sec\theta+tan\theta) ^{2}=\dfrac{csc\theta+1}{csc\theta-1}$

$\huge\tt{\red {To\:\:Prove}}$

★LHS =RHS.

$\huge\tt{\red {Concept\:\:Used:}}$

★By taking LHS, we will try to get it in terms of RHS.

★We will use some basic formulae related to trignometery.

$\huge\tt{\red {Answer:}}$

LHS =

$(sec\theta+tan\theta) ^{2}$

=$\left(\dfrac{1}{cos\theta}+\dfrac{sin\theta}{cos\theta}\right) ^{2}$

=$\left(\dfrac{1+sin\theta}{cos\theta}\right) ^{2}$

=$\dfrac{(1+sin\theta) ^{2}}{cos\theta^{2}}$

=$\dfrac{(1+sin\theta)(1+sin\theta)}{1-sin^{2}\theta}$

=$\dfrac{(1+sin\theta)(1+sin\theta)} {(1+sin\theta) (1-sin\theta) }$

=$\dfrac{\cancel{(1+sin\theta)}(1+sin\theta)} {\cancel{(1+sin\theta)} (1-sin\theta) }$

=$\dfrac{1+sin\theta}{1-sin\theta}$

=$\dfrac{1+\frac{1}{csc\theta}}{1-\frac{1}{csc\theta}}$

=$\dfrac{\frac{1+csc\theta}{csc\theta}}{\frac{1-csc\theta}{csc\theta}}$

=$\dfrac{\frac{1+csc\theta}{\cancel{csc\theta}}}{\frac{1-csc\theta}{\cancel{csc\theta}}}$

=${\underline{\boxed{\red{\dfrac{1+csc\theta}{1-csc\theta}}}}}$

=RHS.

$\large\orange{\boxed{\red{.\degree.\:(sec\theta+tan\theta) ^{2}}=\purple{\dfrac{csc\theta+1}{csc\theta-1}}}}$

Hence proved.

$<marquee scrollamount="1300">♥Answer by Rishabh♥</marquee>$