NCERT Class10 Trigonometry Exercises Solutions,PDFs,Questions,notes

# Trigonometry

## Trigonometry Formulae

$***sin^2θ+cos^2θ=1\\ \; \\ sin^2θ=1-cos^2θ\\\; \\ sinθ=\sqrt{(1+cosθ)(1-cosθ)}\\\; \\ cos^2θ=1-cos^2θ\\\; \\ cosθ=\sqrt{(1+sinθ)(1-sinθ)}\\\; \\ sec^2θ-tan^2θ=1\\\; \\ cosec^2θ-cot^2θ=1\\\; \\ sinθ=cos(90-θ)\\\; \\ sin37=cos(90-37)=cos53\\\; \\ cosθ=sin(90-θ)\\\; \\ cos22=sin68\\\; \\ tanθ=cot(90-θ)\\\; \\ tan41=cot49\\\; \\ cotθ=tan(90-θ)\\\; \\ cot56=tan34\\\; \\ secθ=cosec(90-θ)\\\; \\ sec76=cosec14\\\; \\ cosecθ=sec(90-θ)\\\; \\ cosec65=sec25\\\; \\ If \; \; \\secθ-tanθ=p \; \; then \; \; secθ+tanθ=…..\\\; \\ sec^2θ-tan^2θ=1\; \\ a^2-b^2=(a+b)(a-b)\\\; \\ therefore \; \; sec^2θ-tan^2θ=(secθ+tanθ)(secθ-tanθ)\\ \; \\ So \; \; secθ+tanθ=\frac{sec^2θ-tan^2θ}{(secθ-tanθ)}\\\; \\ secθ+tanθ=\frac{1}{(secθ-tanθ)}\\\; \\ and \;So\; secθ-tanθ=\frac{sec^2θ-tan^2θ}{(secθ+tanθ)}\\\; \\ secθ-tanθ=\frac{1}{(secθ+tanθ)}$

If A+B+C=180 then Show that Sin($\frac{A+B}{2}$)=CoS( $\frac{C}{2}$ )

A+B=180-C —> ($\frac{A+B}{2}$)=90-$\frac{C}{2}$

applying Sin on both sides

Sin($\frac{A+B}{2}$ )=sin(90-$\frac{C}{2}$)

Sin($\frac{A+B}{2}$ )=Cos($\frac{C}{2}$)

If A+B+C=180 then Show that tan($\frac{A+B}{2}$)=cot( $\frac{C}{2}$ )

A+B=180-C —> ($\frac{A+B}{2}$)=90-$\frac{C}{2}$

applying tan on both sides

tan($\frac{A+B}{2}$ )=tan(90-$\frac{C}{2}$)

tan($\frac{A+B}{2}$ )=cot($\frac{C}{2}$)