Question

prove that:
1+sec a/sec a=sin^2a/1+cos a​

Answer 1

Answer:

here is your answer hope u understand

Answer 2

correct Question :-

$\tt\:\dfrac{1+secA}{secA}=\dfrac{{sin}^{2}A}{1-cosA}$

[Hint: simplify LHS and RHS separately]

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$\tt\:LHS=\dfrac{1+secA}{secA}$

$\tt\longrightarrow\:\dfrac{1}{secA}+\cancel\dfrac{secA}{secA}$

$\tt\longrightarrow\:cosA+1$

$\tt\longrightarrow\:1+cosA$

$\tt\:RHS=\dfrac{{sin}^{2}A}{1-cosA}$

$\tt\longrightarrow\:\dfrac{1-{cos}^{2}A}{1-cosA}$

$\tt\longrightarrow\:\dfrac{{(1)}^{2}-{(cosA)}^{2}}{(1-cosA)}$

$\tt\longrightarrow\:\dfrac{\cancel{(1-cosA)}(1+cosA)}{\cancel{(1-cosA)}}$

$\tt\longrightarrow\:1+cosA$

$\tt\:LHS=RHS$

Some Trigonometry formula :-

• sinθ . cosecθ = 1

• cosθ . secθ = 1

• tanθ . cotθ = 1

• sin²θ + cos²θ = 1

• sec²θ - tan²θ = 1

• cosec² - cot² = 1