Question

\(\frac{{sin{{27}^0}cos{{63}^0}}}{{co{s^2}{{27}^0}}} - \frac{{sec{{27}^0}cosec{{63}^0}}}{{ta{n^2}{{45}^0}}}\) is:
1. 2
2. 1
3. 0
4. -1

Answer 1

Correct Answer - Option 4 : -1

Formula used:

sinθ = cos (90 – θ)

secθ = cosec (90 – θ)

sinθ/cosθ = tanθ 

tan2θ – sec2θ = -1

tan²45 = 1

Calculation:

\(\frac{{sin{{27}^0}cos{{63}^0}}}{{co{s^2}{{27}^0}}} - \frac{{sec{{27}^0}cosec{{63}^0}}}{{ta{n^2}{{45}^0}}}\)

⇒ (sinθ × sinθ)/cos²θ – (secθ × secθ)/tan²45

⇒ sin²θ/cos²θ – sec²θ/tan²45

⇒ tan²θ – sec²θ/1

⇒ tan²θ – sec²θ = -1

∴ The required value is -1