Question

solve it!!

no Spamming ❌​

Answer 1

$\tt To\:prove : \frac{1+sinB}{cosB} + \frac{cosB}{1+sinB} = 2secB \\ \\ \tt Taking\:L.H.S. \\ \\ \tt \implies \frac{1+sinB}{cosB} + \frac{cosB}{1+sinB} \\ \\ \tt Cross\: multiplying \\ \\ \tt \implies \frac{{(1+sinB)}^{2} + {cos}^{2}B}{cosB(1+sinB)} \\ \\ \tt \implies \frac{1 + 2sinB + {sin}^{2}B + {cos}^{2}B}{cosB(1+sinB)} \\ \\ \tt \implies \frac{1+2sinB+\green{{sin}^{2}B +{cos}^{2}B}}{cosB(1+sinB)} \\ \\ \tt \implies \frac{1+2sinB +1}{cosB(1+sinB)} \\ \\ \tt \implies \frac{2+2sinB}{cosB(1+sinB)} \\ \\ \tt \implies \frac{2(1+sinB)}{cosB(1+sinB)} \\ \\ \tt \implies \frac{2 \cancel{\orange{\tt (1+sinB)}}}{cosB \cancel{\orange{\tt (1+sinB)}}} \\ \\ \tt \implies \frac{2}{cosB} \\ \\ \tt \implies 2secB \\ \\ \tt \implies R.H.S. \\ \\ \bold{\underline{\red{\tt Hence\:proved }}} \\ \\ \large{\purple{\mathfrak{Formulas\:and\: identities}}} \\ \\ \orange{\mathbb{ IDENTITIES }} \\ \\ \tt \circ \:\: {sin}^{2}\theta + {cos}^{2}\theta = 1 \\ \tt \circ \:\: 1+{tan}^{2}\theta = {sec}^{2}\theta \\ \tt \circ \:\: 1 + {cot}^{2}\theta = {cosec}^{2}\theta \\ \\ \orange{\mathbb{ DOUBLE\:ANGLE\:FORMULAS}} \\ \\ \tt \circ \:\: sin2\theta = 2sin\theta cos\theta \\ \tt \circ \:\: cos2\theta = {cos}^{2}\theta - {sin}^{2}\theta \\ \tt \circ \:\: tan2\theta = \frac{2tan\theta}{1-{tan}^{2}\theta} \\ \\ \orange{\mathbb{TRIPLE\:ANGLE\:FORMULAS}} \\ \\ \tt \circ \:\: sin3\theta = 3sin\theta - 4{sin}^{3}\theta \\ \tt \circ \:\: cos3\theta = 4{cos}^{3}\theta - 3cos\theta \\ \tt \circ \:\: tan3\theta = \frac{3tan\theta - {tan}^{3}\theta}{1-3{tan}^{2}\theta} \\ \\ \orange{\mathbb{GENERAL\: TRANSFORMATIONS}} \\ \\ \tt \circ \:\: cosec\theta = \frac{1}{sin\theta} \\ \tt \circ \:\: sec\theta = \frac{1}{cos\theta} \\ \tt \circ \:\: cot\theta = \frac{1}{tan\theta}$

Answer 2

Question:-

Prove that:-

• $\sf{\dfrac{1 + sinB}{cosB} + \dfrac{cosB}{1 + sinB} = 2 secB}$

Solution:-

Taking LHS,

$\sf{\dfrac{1 + sinB}{cosB} + \dfrac{cosB}{1 + sinB}}$

Taking LCM,

$= \sf{\dfrac{(1 + sinB)^2 + (cosB)^2}{cosB(1 + sinB)}}$

$= \sf{\dfrac{1 + 2sinB + sin^2B + cos^2B}{cosB(1 + sinB)}}$

$= \sf{\dfrac{1 + 2sinB + 1}{cosB(1 + sinB)}}$

$= \sf{\dfrac{2 + 2sinB}{cosB(1 + sinB)}}$

$= \sf{\dfrac{2(1 + sinB)}{cosB(1 + sinB)}}$

$= \sf{\dfrac{2\cancel{(1 + sinB)}}{cosB\cancel{(1 + sinB)}}}$

$= \sf{\dfrac{2}{cosB}}$

$= \sf{2secB}$

Taking RHS,

$\sf{2secB}$

$\sf{\therefore\:LHS = RHS\:[Proved]}$

______________________________________

Identity Used:-

• $\dag{\boxed{\red{\underline{\blue{\sf{\dfrac{1}{cos\theta} = sec\theta}}}}}}$

______________________________________

Some other identities to remember:-

$\boxed{\begin{array}{ccc} \sf{sin\theta} & \leftrightharpoons & \sf{\dfrac{1}{cosec\theta}} \\\\ \sf{cos\theta} & \leftrightharpoons & \sf{\dfrac{1}{sec\theta}} \\\\ \sf{tan\theta} & \leftrightharpoons & \sf{\dfrac{1}{cot\theta}} \end{array}}$

______________________________________