Question

Please answer this question​

Answer 1

Step-by-step explanation:

Correct option is

B

k

2

+1

2k

Given secθ+tanθ=k

As we know that

sec

2

θ−tan

2

θ=1

⟹(secθ+tanθ)(secθ−tanθ)=1

⟹secθ−tanθ=

k

1

⟹secθ=

2

k+

k

1

=

2k

1+k

2

⟹cosθ=

k

2

+1

2k

Answer 2

$\mathbb{LHS ↓} \\\\ \tt \dfrac{k²-1}{k²+1} \\ \\ \tt\dfrac{(tanA + secA)²-1}{(tanA+secA)²+1} \\\\\tt\dfrac{tan²A + sec²A + 2tanAsecA-1}{tan²A+sec²A +2tanAsecA +1} \\ \\ \tt\dfrac{ tan²A+(sec²A-1) +2tanAsecA }{ (1+tan²A) + sec²A + 2tanAsecA } \\\\ \tt\dfrac{ 2tan²A +2tanAsecA}{ 2sec²A + 2tanAsecA }\\ \\ \tt\dfrac{2tanA(secA + tanA)}{2secA(tanA +secA)} \\ \tt\dfrac{tanA}{secA} \\ \\ \tt{cosA}×\dfrac{sinA}{cosA} \\ \\ \tt{sinA} \\\\ \mathbb{RHS ↑}$