Question

Given that tanA + secA = k, show that $\frac{k²+1}{k²-1}$ = cosecA .

Answer 1

Step-by-step explanation:

LHS ↓

$\tt \dfrac{k²+1}{k²-1}$

$\tt\dfrac{(tanA + secA)²+1)}{(tanA+secA)²-1}$

$\tt\dfrac{tan²A + sec²A + 2tanAsecA+1}{tan²A+sec²A +2tanAsecA -1}$

$\tt\dfrac{(1+tan²A) + sec²A + 2tanAsecA}{tan²A+(sec²A-1) +2tanAsecA}$

$\tt\dfrac{2sec²A + 2tanAsecA}{2tan²A +2tanAsecA}$

$\tt\dfrac{2secA(secA + tanA)}{2tanA(tanA +secA)}$

$\tt\dfrac{secA}{tanA}$

$\tt\dfrac{1}{cosA}×\dfrac{cosA}{sinA}$

$\tt\dfrac{1}{sinA}$

$\tt{cosecAcosecA}$

RHS ↑

Answer 2

Answer:

k²−1

k²+1

\tt\dfrac{(tanA + secA)²+1)}{(tanA+secA)²-1}

(tanA+secA)²−1

(tanA+secA)²+1)

\tt\dfrac{tan²A + sec²A + 2tanAsecA+1}{tan²A+sec²A +2tanAsecA -1}

tan²A+sec²A+2tanAsecA−1

tan²A+sec²A+2tanAsecA+1

\tt\dfrac{(1+tan²A) + sec²A + 2tanAsecA}{tan²A+(sec²A-1) +2tanAsecA}

tan²A+(sec²A−1)+2tanAsecA

(1+tan²A)+sec²A+2tanAsecA

\tt\dfrac{2sec²A + 2tanAsecA}{2tan²A +2tanAsecA}

2tan²A+2tanAsecA

2sec²A+2tanAsecA

\tt\dfrac{2secA(secA + tanA)}{2tanA(tanA +secA)}

2tanA(tanA+secA)

2secA(secA+tanA)

\tt\dfrac{secA}{tanA}

tanA

secA

\tt\dfrac{1}{cosA}×\dfrac{cosA}{sinA}

cosA

1

×

sinA

cosA

\tt\dfrac{1}{sinA}

sinA

1

\tt{cosecAcosecA}