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Prove that sinA-cosA+1÷sinA+cosA-1 = 1÷secA-tanA

To prove: \(\begin{array}{l}\frac{sinA-cosA+1}{sinA+cosA-1}=\frac{1}{secA-tanA}\end{array} \) Solution: \(\begin{array}{l}LHS=\frac{sinA-cosA+1}{sinA+cosA-1}\end{array} \) Divinding by cosA  \(\begin{array}{l}\frac{tanA+secA-1}{tanA-secA+1}\end{array} \) =\(\begin{array}{l}\frac{(secA+tanA)-(sec^2A-tan^2A)}{tanA-secA+1}\end{array} \) =\(\begin{array}{l}\frac{(secA+tanA)-(secA-tanA)(secA+tanA)}{tanA-secA+1}\end{array} \) =\(\begin{array}{l}\frac{(secA+tanA)(1-secA+tanA)}{1-secA+tanA}\end{array} \) =secA+tanA =\(\begin{array}{l}\frac{1}{secA-tanA}\end{array}... View Article

A parallel beam of monochromatic light of wavelength 5000 A is incident on a single narrow slit of width 0.001 mm. The light is focussed by a convex lens on a screen placed on a focal plane. The first minimum will be formed for the angle of diffraction equal to a) 50 degrees b) 15 degrees c) 20 degrees d) 30 degrees

Option d) 30 degrees \(\begin{array}{l}dsin\Theta =m\lambda\end{array} \) \(\begin{array}{l}\Theta = sin^{-1}(\frac{m\lambda }{d})\end{array} \) \(\begin{array}{l}sin^{-1}=\frac{1*5000*10^{-10}}{0.001*10^{-3}}\end{array} \) \(\begin{array}{l}\theta =30^{0}\end{array} \)