if A+B=90 DEGREE. SHOW THAT 1+COT2A=SEC2B
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Consider ΔPQR and ΔPXY
∠PQR=∠PXY ( corresponding angles and XY is parallel to QR)
∠QPR= ∠XPY (common)
Therefore ΔPQR∼ΔPXY
This implies
∠PQR=∠PXY ( corresponding angles and XY is parallel to QR)
∠QPR= ∠XPY (common)
Therefore ΔPQR∼ΔPXY
This implies
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