Simplify the expression: tan θ − sec θ + 1tan θ + sec θ − 1.
व्यंजक (Expression) को सरल कीजिए: tan θ − sec θ + 1tan θ + sec θ − 1
Answer और ExplanationAnswer and explanation
Answer: (D) 1 − sin θ/cos θ
हर (Denominator) में 1 की जगह sec²θ − tan²θ लिखें:
tan θ + sec θ − (sec²θ − tan²θ) = (sec θ + tan θ) − (sec θ + tan θ)(sec θ − tan θ)
= (sec θ + tan θ)(1 − sec θ + tan θ)
अंश (Numerator) tan θ − sec θ + 1 वही कोष्ठक (Bracket) (1 − sec θ + tan θ) है।
इसलिए व्यंजक (Expression) = 1sec θ + tan θ = sec θ − tan θ = 1 − sin θcos θ
✓
उत्तर 1 − sin θcos θ है।
Write 1 = sec²θ − tan²θ in the denominator:
tan θ + sec θ − (sec²θ − tan²θ) = (sec θ + tan θ) − (sec θ + tan θ)(sec θ − tan θ)
= (sec θ + tan θ)(1 − sec θ + tan θ)
The numerator tan θ − sec θ + 1 is the same bracket (1 − sec θ + tan θ).
So the expression = 1sec θ + tan θ = sec θ − tan θ = 1 − sin θcos θ
✓
The answer is 1 − sin θcos θ.
SMART APPROACH
θ = 30° रखें (tan θ = 1√3, sec θ = 2√3): 1 − 1√3√3 − 1 = 1√3। विकल्प (D): 1 − 12√32 = 1√3। इसलिए 1 − sin θcos θ।
Put θ = 30° (tan θ = 1√3, sec θ = 2√3): 1 − 1√3√3 − 1 = 1√3. Option (D): 1 − 12√32 = 1√3. So 1 − sin θcos θ.
Source: Selection Post Phase XIV (Graduate) · 22 Sep 2026 · Shift 1। Solution: Prashant Sir, Vijayastra।Source: Selection Post Phase XIV (Graduate) · 22 Sep 2026 · Shift 1. Solution by Prashant Sir, from Vijayastra.