Question

If tan A =2 evaluate sec A sinA +tan²A- cosecA

Answer 1

Answer:

Step-by-step explanation:

tanA=2

∵, sec²A-tan²A=1

or, sec²A=1+tan²A

or, sec²A=1+4

or, secA=√5

∴, cosA=1/secA=1/√5

Again, sin²A+cos²A=1

or, sin²A=1-cos²A

or, sin²A=1-1/5

or, sin²A=4/5

or, sinA=2/√5

∴, cosecA=1/sinA=√5/2

∴, secA.sinA+tan²A-cosecA

=√5.2/√5+(2)²-√5/2

=2+4-√5/2

=6-√5/2

Answer 2

$\bigstar$ Given :

• Tan A = 2

$\bigstar$ To Evaluate :

• Sec A . Sin A + Tan² A - Cosec A

$\bigstar$ Evaluation :

Given Tan A = 2

We know that ,

$\bold{TanA=\frac{opp.side}{adj.side} }\\\\\implies \bold{2=\frac{opp.side}{adj.side} }\\\\\implies \bold{\frac{2}{1}= \frac{opp.side}{adj.side}}$

So , opp. side = 2 , adj.side = 1

Now apply Pythagoras Theorem for finding the hypotenuse.

⇒ opp.side² + adj.side² = hyp.²

⇒ 2² + 1² = hyp.²

⇒ hyp. = √5

$\bold{Now,SecA=\frac{hyp.}{adj.side}=\frac{\sqrt{5}}{1} }\\\\\implies \bold{SecA=\sqrt{5}}\\\\\bold{SinA=\frac{opp.side}{hyp.} =\frac{2}{\sqrt{5}} }\\\\\implies \bold{SinA=\frac{2}{\sqrt{5}} }\\\\\bold{TanA=\frac{opp.side}{adj.side}=\frac{2}{1}}\\\\\implies \bold{TanA=2}\\\\\bold{CosecA=\frac{hyp.}{opp.side}=\frac{\sqrt{5}}{2}}\\\\\implies \bold{CosecA=\frac{\sqrt{5}}{2}}$

Now , our required ,

$\implies \bold{SecA.SinA+Tan^2A-CosecA}\\\\\implies \bold{\sqrt{5}.\frac{2}{\sqrt{5}}+2^2-\frac{\sqrt{5}}{2}}\\\\\implies \bold{2+4-\frac{\sqrt{5}}{2} }\\\\\implies \bold{\bf{\red{\frac{(12-\sqrt{5})}{2}}}}$