Prove this Trigonometric Identity | Step-by-Step Explanation

#OlympiadMathematics #OlympiadPreparation #CollegeEntranceExam

[HappyFamilyOnline]

Very easy step by step explanation👍
Thinks for sharing😊😊

[ikeetkroketjes]

(tanθ+secθ-1)/(tanθ-secθ+1)=
(1+sinθ)/cosθ
in a right triangle tanα=a/b, cosα=b/c, sinα=a/c, secα=c/b so substitute

now take a look at the left hand side, multiply by b/b to remove the ratio’s
(a+c-b)/(a+b-c)
same on the right side
(c+a)/b
so (a-b+c)/(a+b-c)=(a+c)/b
cross multiply
ab-b^2+bc=a^2+ab+bc-c^2
cancel everyting out
-b^2=a^2-c^2
a^2+b^2=c^2 and that is proven by pythagoras

[rahulpaul]

Very nicely solved this I'm more appreciate

[akshaybisen]

You brought back my memories of 10th standard. It was my weak zone. but now, things has been changed. It was year 2013 when i was in 10th and it was really hard to me. Thank you.

[surajkumardas]

Very useful for me this trigonometric related video ❣️

[rifatshan]

To be honest, i don't know trigonometry, but sir this is nice again 😊

[guyhoghton]

This one is a good exercise for revising trig identities, but it can be done a little more simply using proof by contradiction.
First, eliminate values of θ which are multiples of π/2 as the left hand side is undefined for these.
Suppose (tanθ + secθ - 1)/(tanθ - secθ + 1) ≠ (1 + sinθ)/cosθ for some value of θ ≠ nπ/2, n = 0, ±1, ±2, ...
∴ cosθ (tanθ + secθ - 1) ≠ (1 + sinθ)(tanθ - secθ + 1)
∴ cosθtanθ + cosθsecθ - cosθ ≠ tanθ - secθ + 1 + sinθtanθ - sinθsecθ + sinθ
∴ sinθ + 1 - cosθ ≠ tanθ - secθ + 1 + sin²θ/cosθ - tanθ + sinθ
= 1 - (1 - sin²θ)/cosθ + sinθ
= 1 - cos²θ/cosθ + sinθ
∴ sinθ + 1 - cosθ ≠ sinθ + 1 - cosθ, a contradiction.

[nirupamasingh]

Sir this is an important and typical question. V systematically solved. V nice

[burlfromlondon]

Perhaps an alternative method would have been to start by multiplying each & every term by Cosine Theta?

[math]

very well done bro, thanks for sharing this trig identity proof

[chinrxhh]

Like your voice.✌️✌️✌️Clear explanation

[williamwingo]

Here's a different method, which took more time than you want to know:

We have (tanθ + secθ − 1)/(tanθ − secθ + 1); since tanθ = sinθ/cosθ; secθ = 1/cosθ; and 1 = cosθ/cosθ, we can substitute and get:
(sinθ/cosθ + 1/cosθ − cosθ/cosθ)/(sinθ/cosθ − 1/cosθ + cosθ/cosθ); all the interior denominators are equal to cosθ; multiplying top and bottom by cosθ:
(sinθ + 1 − cosθ)/(sinθ − 1 + cosθ); collecting terms:
[sinθ + (1 − cosθ)]/[sinθ − (1 − cosθ)]; now let Q = (1 − cosθ); substituting, we have:
(sinθ + Q)/(sinθ − Q); multiplying top and bottom by (sinθ + Q):
(sinθ + Q)^2/[(sinθ − Q)(sinθ + Q)]; applying the rule in the denominator:
(sinθ + Q)^2/[(sin^2(θ) − Q^2)]; multiplying out Q^2:
(sin^2(θ) + Q^2 +2Qsinθ)/[(sin^2(θ) − Q^2)]; since Q = (1 − cosθ), then Q^2 = 1 + cos^2(θ) − 2cosθ; substituting, we get:
(sin^2(θ) + 1 + cos^2(θ) − 2cosθ +2(1 − cosθ)sinθ)/[(sin^2(θ) − (1 + cos^2(θ) − 2cosθ)]; removing some parentheses and rearranging:
(sin^2(θ) + cos^2(θ) + 1 − 2cosθ +2sinθ(1 − cosθ))/[(sin^2(θ) − cos^2(θ) − 1 + 2cosθ)]; remembering that sin^2(θ) + cos^2(θ) = 1 and therefore sin^2(θ) − cos^2(θ) = 1− 2cos^2(θ):
(1 + 1 − 2cosθ +2sinθ (1 − cosθ))/[(1− 2cos^2(θ) − 1 + 2cosθ)]; collecting terms:
(2 − 2cosθ +2sinθ (1 − cosθ))/[(1− 1 − 2cos^2(θ) + 2cosθ)]; rearranging a little:
(2 − 2cosθ +2sinθ (1 − cosθ))/[2cosθ − 2cos^2(θ)]; dividing numerator and denominator through by 2:
(1 − cosθ + sinθ (1 − cosθ))/[(cosθ − cos^2(θ)]; factoring out (cosθ) in the denominator and collecting terms in the numerator:
[(1 − cosθ)(1 + sinθ)]/[cosθ(1 − cosθ)]; and finally, dividing top and bottom by (1 − cosθ), assuming it's not zero:
(1 + sinθ)/cosθ: QED; Whew!
I was sure I could get it to go this way, but it took several trips through before I got all the signs right. Good thing I'm dedicated.🤠

[navindraarjunaabeyesekera]

How can you cancel out (1-Sec theta +tan theta ) and ( Tan theta -Sec theta + 1) ???

[mustafizrahman]

Sir, it came in our exam. Even this is our textbook question. I solved then it easily.

[giuseppemalaguti]

Basta moltiplicare il D1*N2=D2*N1 e l'identità é provata automaticamente

[susennath]

Again nostralgic

Class ten of the year 1991.

Can you feel my feelings?

[hoitinleung]

identity invalid when the angle is zero

[nirupamasingh]

Sir this is an important and typical question. V systematically solved. V nice