Question

15 Show that tan 15° + tan 75°
sec^2 15°/✓sec^2 15° -1​

Answer 1

Answer:

that's 1-1=,0

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Answer 2

Step-by-step explanation:

To prove : tan15°+ tan75° = $\frac{sec^215}{\sqrt{sec^215-1} }$

LHS = tan15°+ tan75°

       = $\frac{sin15}{cos15}$ + $\frac{sin75}{cos75}$

       = $\frac{sin15.cos75+sin75.cos15}{cos15.cos75}$

       = $\frac{sin(15+75)}{cos15.cos75}$

       = $\frac{sin90}{cos15.cos75}$ = $\frac{1}{cos15.cos75}$

       = $\frac{1}{cos15}$*.$\frac{1}{cos75}$

       = sec15°.sec75°    -------------------- (1)

RHS = $\frac{sec^215}{\sqrt{sec^215-1} }$

        = $\frac{sec^215}{\sqrt{tan^215} }$  = $\frac{sec^215}{tan15}$

        = $\frac{ [\frac{1}{cos^215}] } {[ \frac{sin15}{cos15} ]}$

        = $\frac{1}{cos^215}$ * $\frac{cos15}{sin15}$

        = $\frac{1}{sin15.cos15}$ = $\frac{1}{sin(90-75).cos15}$

        = $\frac{1}{cos15.cos75}$

       = $\frac{1}{cos15}$*.$\frac{1}{cos75}$

       = sec15°.sec75°    -------------------- (2)

so, we see that (1) = (2)

=> LHS = RHS

Hence proved