Question

Prove that:
\[1.\frac {tanA+secA-1}{tanA-secA+1}=\frac{1+sinA}{cosA}\] and
\[2.(tanA+cosecB)^{2}-(cotB-secA)^{2}=2tanAcotB(cosecA+secB)\]

Answer 1

You can substitute:

\[1 = sec^2 \theta - tan^2 \theta\]

Answer 2

In the first one...

Answer 3

full sol. pls

Answer 4

Or you can say:

\[1 = sec^2 \theta - tan^2 \theta = (sec \theta + tan \theta)(sec \theta - tan \theta)\]

Try to substitute it there and show me what you got..

Answer 5

i will do it on paper.. wait a moment

Answer 6

\[\frac{(\tan A + secA) (tanA - secA +1)}{(tanA - secA +1)} \implies tanA + \sec A\]

Answer 7

As waterineyes said,
put 1 = (secA+tanA)(secA-tanA) in denominator..
we will evaluate denominator first.
D = tanA-secA+(tanA+secA)(secA-tanA)
take (secA-tanA) common
(secA-tanA)[tanA+secA-1]

now numerator = tanA+secA-1.. numerator and denominator will cancel tanA+secA-1..
which leaves expression as
1
--------- ...
secA-tanA

we know that (secA+tanA)(secA-tanA)=1
=> secA+tanA = 1/(secA-tanA)

this makes the above expression = secA+tanA
=(1/cosA) + (sinA/cosA)
=(1+sinA)/cosA

I tried my best lol.. :) it takes too much time to type.. :P so try 2nd one yourself

Answer 8

\[tanA + secA = \frac{sinA}{cosA} + \frac{1}{cosA} = \frac{1 + sinA}{cosA}\]

Answer 9

lol ^^ nice one :P

Answer 10

should have done this

Answer 11

@shubham.bagrecha u there bro ?

Answer 12

\[\frac{tanA + secA - (tanA + secA)(tanA - secA)}{tanA - secA + 1}\]

\[\frac{(tanA + secA)(1 - secA + \tan A)}{\tan A - \sec A + 1} \implies \frac{(tanA + secA)(\tan A - \sec A + 1)}{\tan A - \sec A + 1}\]

\[\implies tanA + secA \implies \frac{1 + sinA}{cosA}\]

Answer 13

He is only here for full solutions..

Answer 14

I remember there was same question like this in my final exams :P

Answer 15

now 2nd ques.

Answer 16

More precisely if I can guess It was in CBSE 2003 In India..

Answer 17

I gave my 10th standard paper in March 2012 lol

Answer 18

anyways i am attempting 2nd one

Answer 19

Ha ha ha..

Answer 20

hey i am stuck .. hehe

Answer 21

Very nice question..
Love to do the second one..

Answer 22

In second the identities using are:

\[\sec^2A - \tan^2 A = 1\]

\[cosec^2 A - \cot^2 A = 1\]

\[cotB = \frac{cosB}{sinB}\]

\[tanA = \frac{sinA}{cosA}\]

Answer 23

So,

firstly open the brackets by applying the squares to the brackets:

\[\tan^2A + cosec^2 A + 2tanA \cdot cosecA - \cot^2B - \sec^2B +2cotB\cdot secB\]
After applying the first two identities: 1 and -1 will cancel out:

\[\implies 2(tanA cosecA + cotBsecB)\]

Take \(tanAcotB\) common:

\[\implies 2tanA cotB(\frac{cosecB}{cotB} + \frac{secA}{tanA})\]

\[\implies 2tanA cotB(cosecA + secB)\]

Answer 24

ok thanks

Answer 25

Second last step I simplified as:

\[\frac{cosecB}{cotB} + \frac{secA}{tanA} \implies \frac{cosecB \cdot sinB}{cosB} + \frac{secA \cdot cosA}{sinA}\]

\[\implies \frac{1}{cosB} + \frac{1}{sinA} \implies (secB + cosecA)\]

Answer 26

Welcome dear..