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प्रश्न
Prove that `(sin (90° - θ))/cos θ + (tan (90° - θ))/cot θ + (cosec (90° - θ))/sec θ = 3`.
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उत्तर
LHS = `(sin (90° - θ))/cos θ + (tan (90° - θ))/cot θ + (cosec (90° - θ))/sec θ `
= `cos θ/cos θ + cot θ/cot θ + sec θ/sec θ`
= 1 + 1 + 1
= 3
= RHS
Hence proved.
संबंधित प्रश्न
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