Question

tana+seca-1/tana-seca+1 = 1+sina/cosa​

Answer 1

❍ Topic :-

Trigonometry

❍Given to prove :-

$\dfrac{tanA+secA-1}{tanA-secA+1} = \dfrac{1+sinA}{1-sinA}$

❍SOLUTION:-

We shall take LHS and we prove RHS

LHS = $\dfrac{tanA+secA-1}{tanA-secA+1}$

Here 1 can be written as

sec²A - tan²A = 1 [from Trigonometric Identities]

sec²A - tan²A = (secA+tanA)(secA-tanA)

$\dfrac{tanA+secA-(sec^2A-tan^2A)}{tanA-secA+1}$

$\dfrac{secA+tanA - (secA+tanA)(secA-tanA)}{tanA-secA+1}$

Take common secA + tanA

${secA + tanA} \dfrac{1-(secA-tanA)}{tanA-secA+1}$

${secA + tanA} \dfrac{1-secA+tanA}{1-secA+tanA}$

${secA + tanA (1)}$

${secA + tanA}$

secA = 1/cosA

tanA = sinA/cosA

$\dfrac{1}{cosA} + \dfrac{sinA}{cosA}$

$\dfrac{1+sinA}{cosA}$ = LHS

Since ,

LHS = RHS

PROVED

❍Know more :-

❍Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

❍Trigonometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

❍Trigonometric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj

Answer 2

Proved!

Step-by-step explanation:

We are given a trigonometric equation:

→ (tanA + secA - 1)/(tanA - secA + 1) = (1 + sinA)/cosA

First, let us solve the left hand side of the equation.

→ (tanA + secA - 1)/(tanA - secA + 1)

• [We know that, sec²A - tan²A = 1 (Trigonometric identity). On substituting the identity in 1 at numerator, we get:]

→ [tanA + secA - (sec²A - tan²A)]/(tanA - secA + 1)

• [We know that, (a² - b²) = (a + b)(a - b) (Algebraic identity). Similarly,]

→ [tanA + secA - (secA + tanA)(secA - tanA)]/(tanA - secA + 1)

• [Let us take tanA and secA as common.]

→ tanA + secA[1 - (secA - tanA)]/(tanA - secA + 1)

→ tanA + secA(tanA - secA + 1)/(tanA - secA + 1)

• [tanA - secA + 1 gets cancelled.]

→ tanA + secA

We know that,

• tanA = sinA/cosA
• secA = 1/cosA

Therefore,

→ sinA/cosA + 1/cosA

→ (1 + sinA)/cosA

[LHS = RHS]

Hence, Proved!