Question

prove that cot2(90- theta) / tan 2 theta - 1 + cosec2theta / sec2theta - cos2theta = 1 / sin2theta - cos2theta​

Answer 1

Answer:

Step-by-step explanation:

Answer:

\frac{tan^{2}\theta}{tan^{2}\theta-1}+\frac{cosec^{2}\theta}{sec^{2}\theta-cosec^{2}\theta}=\frac{1}{sin^{2}\theta-cos^{2}\theta}

Step-by-step explanation:

LHS=\frac{tan^{2}\theta}{tan^{2}\theta-1}+\frac{cosec^{2}\theta}{sec^{2}\theta-cosec^{2}\theta}

=\frac{\frac{sin^{2}\theta}{cos^{2}\theta}}{\frac{sin^{2}\theta}{cos^{2}\theta}-1}+\frac{\frac{1}{sin^{2}\theta}}{\frac{1}{cos^{2}\theta}-\frac{1}{sin^{2}\theta}}

=\frac{\frac{sin^{2}\theta}{cos^{2}\theta}}{\frac{sin^{2}\theta-cos^{2}\theta}{cos^{2}\theta}}+\frac{\frac{1}{sin^{2}\theta}}{\frac{sin^{2}\theta-cos^{2}\theta}{cos^{2}\theta sin^{2}\theta}}

=\frac{sin^{2}\theta}{sin^{2}\theta-cos^{2}\theta}+\frac{cos^{2}\theta}{sin^{2}\theta-cos^{2}\theta}

=\frac{sin^{2}\theta+cos^{2}\theta}{sin^{2}\theta-cos^{2}\theta}

=\frac{1}{sin^{2}\theta-cos^{2}\theta}\\=RHS

Therefore,

\frac{tan^{2}\theta}{tan^{2}\theta-1}+\frac{cosec^{2}\theta}{sec^{2}\theta-cosec^{2}\theta}=\frac{1}{sin^{2}\theta-cos^{2}\theta}

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