Question

Prove that (1+cotA - cosecA) (1+tanA+secA) = 2.​

Answer 1

Step-by-step explanation:

(1+cosA/sinA -1/sinA)(1+cosA/sinA + 1/sinA)

(sinA+cosA-1/sinA) (sinA+cosA+1 /cosA)

$[sinA+cosA]^{2}$ -1/sinA*cosA

1+2*sinAcosA -1/sinA*cosA                 [therefore $sin^{2}$A +$cos^{2}$A =1]

=2

Answer 2

${\bold{\pink{\underline{\red{So}\purple{lut}\green{ion}\orange{:-}}}}}$

$\rm\:LHS=(1+cotA-cosecA)(1+tanA+secA)$

$\rm\:=\bigg(1+\dfrac{cosA}{sinA}-\dfrac{1}{sinA}\bigg)\:\bigg(1+\dfrac{sinA}{cosA}+\dfrac{1}{cosA}\bigg)$

$\rm\:=\bigg(\dfrac{sinA+cosA-1}{sinA}\bigg)\:\bigg(\dfrac{cosA+sinA+1}{cosA}\bigg)$

$\rm\:=\dfrac{(sinA+cosA)-1.\:(sinA+cosA)+1}{sinA\:cosA}$

$\rm\:=\dfrac{(sinA+cosA)^2-1}{sinA\:cosA}$

$\rm\:=\dfrac{sin^2{A}+cos^2{A}+2\:sinA\:cosA-1}{sinA\:cosA}$

$\rm\:=\dfrac{1+2\:sinA\:cosA-1}{sinA\:cosA}$

$\rm\:=\dfrac{2\:sinA\:cosA}{sinA\:cosA}=2=RHS$

$\rm\:LHS=RHS$

$\rm\:Hence,proved$

$\rm\boxed\star\purple{\underline{\underline{{More\: Information.....!!!!!!}}}}$

Some trigonometric identities :-

• $\rm\:{sin}^{2}\theta+{cos}^{2}\theta=1$

• $\rm\:1+{tan}^{2}\theta={sec}^{2}\theta$

• $\rm\:1+{cot}^{2}\theta={cosec}^{2}\theta$

• $\rm\:tan\theta=\dfrac{sin\theta}{cos\theta}$

• $\rm\:tan\theta\:\times\:cot\theta=1$